رياضيات فصل أول

 أجد $\frac{dy}{dx}$  لكل مما يأتي:

                 $$\begin{gathered} a) x^2+y^2=13 \\ Solution: \\ 2x+2y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=- \frac{x}{y} \end{gathered}$$         

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        $$\begin{gathered} b) 2x+5y^2=siny \\ Solution: \\ 2+10y\frac{dy}{dx}=cosy\frac{dy}{dx} \\ 10y\frac{dy}{dx}-cosy\frac{dy}{dx}=-2 \\ \frac{dy}{dx}(10y-cosy)=-2 \Rightarrow \frac{dy}{dx}=-\frac{2}{10y-cosy} \end{gathered}$$          

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أجد $\frac{dy}{dx}$ لكل مما يأتي:

          $$\begin{gathered} a) 3x{y}^2+y^3=8 \\ Solution: \\ 3y^2+3x(2y)\frac{dy}{dx}+3y^2\frac{dy}{dx}=0 \\ \frac{dy}{dx}(6xy+3y^2)=-3y^2 \\ \Rightarrow \frac{dy}{dx}=- \frac{3y^2}{6xy+3y^2}=- \frac{y}{2x+y} \end{gathered}$$           

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          $$\begin{gathered} b) tan(x-y)=2x{y}^3+1 \\ Solution: \\ se{c}^2(x-y\left)\right(1-\frac{dy}{dx})=2y^3+6x{y}^2\frac{dy}{dx} \\ se{c}^2(x-y)-\frac{dy}{dx}se{c}^2(x-y)=2y^3+6x{y}^2\frac{dy}{dx} \\ -\frac{dy}{dx}se{c}^2(x-y)-6x{y}^2\frac{dy}{dx}=2y^3-se{c}^2(x-y) \\ -\frac{dy}{dx}(se{c}^2(x-y)+6x{y}^2)=2y^3-se{c}^2(x-y) \\ \frac{dy}{dx}=-\frac{2y^3-se{c}^2(x-y)}{se{c}^2(x-y)+6x{y}^2} \end{gathered}$$            

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                   $$\begin{gathered} c) x^2=\frac{x-y}{x+y} \\ Solution: \\ x^2(x+y)=x-y \\ x^3+ x^{2}y=x-y \\ 3x^2+2xy+x^2\frac{dy}{dx}=1-\frac{dy}{dx} \\ x^2\frac{dy}{dx}+\frac{dy}{dx}=1-3x^2-2xy \\ \frac{dy}{dx}( x^2+1)=1-3x^2-2xy \\ \frac{dy}{dx}=\frac{1-3x^2-2xy}{x^2+1} \end{gathered}$$           

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a)   أجد ميل مماس منحنى العلاقة: $y^2=lnx$عند النقطة $(e , 1)$.

           $$\begin{gathered} Solution: \\ y^2=lnx ,x_1=e ,y_1=1 , m=\frac{dy}{dx} \\ 2y\frac{dy}{dx}=\frac{1}{x} \\ \frac{dy}{dx}=\frac{1}{2yx}= \frac{1}{2\left(1\right)\left(e\right)}=\frac{1}{2e} \end{gathered}$$           

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b)   أجد ميل مماس منحنى العلاقة: ${(y-3)}^2=4(x-5)$عند $x = 6$  .

                $$\begin{gathered} Solution: \\ {(y-3)}^2=4(x-5) \\ 2(y-3)\frac{dy}{dx}=4 \Rightarrow \frac{dy}{dx}=\frac{2}{y-3} \\ when x=6 \Rightarrow {(y-3)}^2=4(6-5) \\ \Rightarrow {(y-3)}^2=4 \\ \Rightarrow y-3=2 \Rightarrow y=5 \\ or y-3=-2 \Rightarrow y=1 \\ \frac{dy}{dx}=\frac{2}{5-3}=1 when y=5 \\ \frac{dy}{dx}=\frac{2}{1-3}=-1 when y=1 \end{gathered}$$           

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أجد معادلة المماس لمنحنى العلاقة : $x^3+y^3-3xy=17$ عند النقطة $(2 , 3)$.  

     $$\begin{gathered} Solution: \\ x^3+y^3-3xy=17 \\ m=\frac{dy}{dx} \\ 3x^2+3y^2\frac{dy}{dx}-3y-3x\frac{dy}{dx}=0 when x=2 , y=3 \\ 3{\left(2\right)}^2+3{\left(3\right)}^2\frac{dy}{dx}-3\left(3\right)-3\left(2\right)\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{1}{7} \\ y-y_1=m(x-x_1) \\ y-3=-\frac{1}{7}(x-2) \end{gathered}$$           

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إذا كان $xy+y^2=2x$ ، فأجد $\frac{d^{2}y}{d{x}^2}$ .

        $$\begin{gathered} Solution: \\ xy+y^2=2x \\ y+x\frac{dy}{dx}+2y\frac{dy}{dx}=2 \\ \frac{dy}{dx}(x+2y)=2-y \\ \frac{dy}{dx}=\frac{2-y}{x+2y} \\ \frac{d^{2}y}{d{x}^2}=\frac{-\frac{dy}{dx}\times (x+2y)-(2-y)(1+2\frac{dy}{dx})}{{(x+2y)}^2} \\ =\frac{-\left(\frac{2-y}{x+2y}\right)\times (x+2y)-(2-y)(1+2\left(\frac{2-y}{x+2y}\right))}{{(x+2y)}^2} \\ =\frac{-(2-y)(x+2y)-(2-y)(1+2(2-y))}{{(x+2y)}^3} \\ \frac{d^{2}y}{d{x}^2}=-\frac{(2-y)(x+5)}{{(x+2y)}^3} \end{gathered}$$          

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       أجد $\frac{d^{2}y}{d{x}^2}$ للمعادلة الوسيطية الآتية عندما   . t = 2

                     $x=3t^2+1 , y=t^3-2t^2$

           $$\begin{gathered} Solution: \\ x=3t^2+1 \Rightarrow \frac{dx}{dt}=6t \\ y=t^3-2t^2 \Rightarrow \frac{dy}{dt}=3t^2-4t \\ \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=(3t^2-4t)\times \frac{1}{6t}=\frac{1}{2}t-\frac{2}{3} \\ \frac{dy}{dx}=\frac{1}{2}t-\frac{2}{3} \\ \frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dt}(\frac{1}{2}t-\frac{2}{3}) \\ \frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{1}{2} \\ \frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{\frac{1}{2}}{6t}=\frac{1}{12t} \\ when t=2 \\ \frac{d^{2}y}{d{x}^2}=\frac{1}{12\left(2\right)}=\frac{1}{24} \end{gathered}$$         

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أجد مشتقة كل اقتران مما يأتي باستعمال الاشتقاق اللوغاريتمي:

$$\begin{gathered} a) y=x^{\sqrt{x} } \\ Solution: \\ ln y=ln x^{\sqrt{x} } \\ ln y=\sqrt{x}ln x^{ } \\ \frac{d}{dx}( ln y)= \frac{d}{dx}(\sqrt{x}ln x) \\ \frac{\frac{dy}{dx}}{y}=\frac{1}{2\sqrt{x}}lnx +\sqrt{x} \times \frac{1}{x} \\ \frac{dy}{dx}=y(\frac{1}{2\sqrt{x}}lnx +\frac{1}{\sqrt{x}}) \\ \frac{dy}{dx} = x^{\sqrt{x} }(\frac{1}{2\sqrt{x}}lnx +\frac{1}{\sqrt{x}}) ; y=x^{\sqrt{x} } \end{gathered}$$          

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$$\begin{gathered} b) y=\sqrt{\frac{x-1}{x^4+1}} \\ Solution: \\ ln y= ln {\left(\frac{x-1}{x^4+1}\right)}^{\frac{1}{2}} \\ ln y= \frac{1}{2} ln \left(\frac{x-1}{x^4+1}\right) \\ ln y= \frac{1}{2}(ln(x-1)-ln(x^4+1)) \\ \frac{d}{dx}( ln y)= \frac{1}{2}\frac{d}{dx}(ln(x-1)-ln(x^4+1)) \\ \frac{\frac{dy}{dx}}{y}= \frac{1}{2}(\frac{1}{x-1}-\frac{4x^3}{x^4+1}) \\ \frac{dy}{dx}= \frac{1}{2}y(\frac{1}{x-1}-\frac{4x^3}{x^4+1}) \\ \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{x-1}{x^4+1}}(\frac{1}{x-1}-\frac{4x^3}{x^4+1}) \end{gathered}$$           

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أجد $\frac{dy}{dx}$  لكل مما يأتي: 

          $$\begin{gathered} 1) x^2-2y^2=4 \\ Solution: \\ 2x-4y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{x}{2y} \end{gathered}$$           

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         $$\begin{gathered} 2) \frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{10} \\ Solution: \\ \frac{-2x}{x^4}+\frac{-2y{\displaystyle \frac{dy}{dx}}}{y^4}=0 \\ \frac{{\displaystyle \frac{dy}{dx}}}{y^3}=\frac{-1}{x^3} \Rightarrow \frac{dy}{dx}=-\frac{y^3}{x^3} \end{gathered}$$            

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          $$\begin{gathered} {3) (x^2+y^2)}^2=50(x^2-y^2) \\ Solution: \\ 2\left(x^2+y^2)\right(2x+2y\frac{dy}{dx})=50(2x-2y\frac{dy}{dx}) \\ (x^2+y^2)(x+y\frac{dy}{dx})=25(x-y\frac{dy}{dx}) \\ x^3+x^{2}y\frac{dy}{dx}+y^{2}x+y^3\frac{dy}{dx}=25x-25y\frac{dy}{dx} \\ x^{2}y\frac{dy}{dx}+y^3\frac{dy}{dx}+25y\frac{dy}{dx}=25x-x^3-y^{2}x \\ \frac{dy}{dx}(x^{2}y+y^3+25y)=25x-x^3-y^{2}x \\ \frac{dy}{dx}=\frac{25x-x^3-y^{2}x}{x^{2}y+y^3+25y} \end{gathered}$$            

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           $$\begin{gathered} 4) e^{x}y=x{e}^y \\ Solution: \\ {e}^{x}y+e^x\frac{dy}{dx}=e^y+x{e}^y\frac{dy}{dx} \\ {e}^x\frac{dy}{dx}-x{e}^y\frac{dy}{dx}=e^y-e^{x}y \\ \frac{dy}{dx}(e^x-x{e}^y)=e^y-e^{x}y \\ \frac{dy}{dx}=\frac{e^y-e^{x}y}{e^x-x{e}^y} \end{gathered}$$           

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            $$\begin{gathered} 5) 3^x=y-2yx \\ Solution: \\ {3}^{x}ln3=\frac{dy}{dx}-2\frac{dy}{dx}x-2y \\ {3}^{x}ln3+2y=\frac{dy}{dx}-2\frac{dy}{dx}x \\ \frac{dy}{dx}(1-2x)=3^{x}ln3+2y \\ \frac{dy}{dx}=\frac{3^{x}ln3+2y}{1-2x} \end{gathered}$$           

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            $$\begin{gathered} 6) \sqrt{x}+\sqrt{y}=5 \\ Solution: \\ \frac{1}{2\sqrt{x}}+\frac{{\displaystyle \frac{dy}{dx}}}{2\sqrt{y}}=0 \\ \frac{{\displaystyle \frac{dy}{dx}}}{2\sqrt{y}}=-\frac{1}{2\sqrt{x}} \\ \frac{dy}{dx}=-\sqrt{\frac{y}{x}} \end{gathered}$$           

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          $$\begin{gathered} 7) x=sec\frac{1}{y} \\ Solution: \\ 1=sec\frac{1}{y}tan\frac{1}{y}(-\frac{{\displaystyle \frac{dy}{dx}}}{y^2}) \\ \frac{dy}{dx}=-\frac{y^2}{sec\frac{1}{y}tan\frac{1}{y}} \end{gathered}$$          

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         $$\begin{gathered} {8) (\sin \pi x +\cos \pi y)}^3=8 \\ Solution: \\ 3{(sin\pi x +cos\pi y)}^2(\pi cos\pi x -\pi \frac{dy}{dx}sin\pi y)=0 \\ (\pi cos\pi x -\pi \frac{dy}{dx}sin\pi y)=\frac{1}{2{(sin\pi x +cos\pi y)}^2} \\ -\pi \frac{dy}{dx}sin\pi y=\frac{1}{3{(sin\pi x +cos\pi y)}^2}-\pi cos\pi x \\ \frac{dy}{dx}=\frac{\pi cos\pi x}{\pi sin\pi y}-\frac{1}{3{(sin\pi x +cos\pi y)}^2\left(\pi sin\pi y\right)} \\ \frac{dy}{dx}=cot\pi x-\frac{1}{3{(sin\pi x +cos\pi y)}^2\left(\pi sin\pi y\right)} \end{gathered}$$           

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         $$\begin{gathered} 9) \frac{x}{y^2}+\frac{y^2}{x}=5 \\ Solution: \\ \frac{x}{y^2}+\frac{y^2}{x}=5 with y^{2}x \\ \Rightarrow x^2+y^4=5y^{2}x \\ 2x+4y^3\frac{dy}{dx}=10y\frac{dy}{dx}x+5y^2 \\ 4y^3\frac{dy}{dx}-10y\frac{dy}{dx}x=5y^2-2x \\ \frac{dy}{dx}(4y^3-10yx)=5y^2-2x \\ \frac{dy}{dx}=\frac{5y^2-2x}{4y^3-10yx} \end{gathered}$$           

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               $$\begin{gathered} 10) x+y=cos(xy) \\ Solution: \\ 1+\frac{dy}{dx}=-sin(xy\left)\right(y+x\frac{dy}{dx}) \\ 1+\frac{dy}{dx}=-y sin(xy)- x\frac{dy}{dx}sin(xy) \\ \frac{dy}{dx}+ x\frac{dy}{dx}sin(xy)=-y sin(xy)-1 \\ \frac{dy}{dx}(1+ xsin\left(xy\right))=-y sin(xy)-1 \\ \frac{dy}{dx}=-\frac{y sin\left(xy\right)+1}{1+ xsin\left(xy\right)} \end{gathered}$$           

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           $$\begin{gathered} 11) x^2+y^2=ln{(x+y)}^2 \\ Solution: \\ {x}^2+y^2=2ln(x+y) \\ 2x +2y\frac{dy}{dx}=2\frac{1+\frac{dy}{dx}}{x+y } \\ {x}^2+xy\frac{dy}{dx}+xy+y^2\frac{dy}{dx}=1+\frac{dy}{dx} \\ xy\frac{dy}{dx}+y^2\frac{dy}{dx}-\frac{dy}{dx}=1-x^2-xy \\ \frac{dy}{dx}(xy+y^2-1)=1-x^2-xy \\ \frac{dy}{dx}=\frac{1-x^2-xy}{xy+y^2-1} \end{gathered}$$            

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           $$\begin{gathered} 12) sinx cosy=x^2-5y \\ Solution: \\ cosx cosy+ sinx(-\frac{dy}{dx}siny)=2x-5\frac{dy}{dx} \\ - sinx(\frac{dy}{dx}siny)+5\frac{dy}{dx}=2x- cosx cosy \\ \frac{dy}{dx}(-sinx siny+5)=2x- cosx cosy \\ \frac{dy}{dx}=\frac{2x- cosx cosy}{5-sinx siny} \end{gathered}$$         

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أجد $\frac{dy}{dx}$ لكلٌّ ممّا يأتي عند القيمة المعطاة:

          $$\begin{gathered} 13) 2y^2+2xy-1=0 , x=\frac{1}{2} \\ Solution: \\ 4y\frac{dy}{dx}+2y+2x\frac{dy}{dx}=0 \\ when x=\frac{1}{2} \\ 2y^2+2(\frac{1}{2})y-1=0 \\ 2y^2+y-1=0 \\ (2y-1\left)\right(y+1)=0 \\ \Rightarrow y=\frac{1}{2} , y=-1 \\ 2y\frac{dy}{dx}+y+x\frac{dy}{dx}=0 when x=\frac{1}{2} , y=\frac{1}{2} \\ 2(\frac{1}{2})\frac{dy}{dx}+(\frac{1}{2})+(\frac{1}{2})\frac{dy}{dx}=0 \\ \frac{3}{2}\frac{dy}{dx}=-\frac{1}{2} \Rightarrow \frac{dy}{dx}=-\frac{1}{3} \\ when x=\frac{1}{2} , y=-1 \\ 2(-1)\frac{dy}{dx}+(-1)+(\frac{1}{2})\frac{dy}{dx}=0 \\ (-\frac{3}{2})\frac{dy}{dx}=1 \Rightarrow \frac{dy}{dx}=-\frac{2}{3} \end{gathered}$$              

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           $$\begin{gathered} 14) y^3+2x^2=11y , y=1 \\ Solution: \\ 3y^2\frac{dy}{dx}+4x=11\frac{dy}{dx} \\ when y=1 \\ {\left(1\right)}^3+2x^2=11(1) \Rightarrow x=\sqrt{5} \\ \Rightarrow 3{\left(1\right)}^2\frac{dy}{dx}+4\sqrt{5}=11\frac{dy}{dx} \\ \Rightarrow 8\frac{dy}{dx}=4\sqrt{5} \Rightarrow \frac{dy}{dx}=\frac{2}{\sqrt{5}} \end{gathered}$$               

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أجد ميل المماس لمنحنى كل علاقة مما يأتي عند النقطة المعطاة

          $$\begin{gathered} 15) x^2+y^2=25 , (3 ,-4) \\ Solution: \\ 2x +2y\frac{dx}{dy}=0 \\ when x=3 , y=-4 \\ 2(3) +2(-4)\frac{dx}{dy}=0 \Rightarrow \frac{dx}{dy}=\frac{3}{4} \end{gathered}$$             

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          $$\begin{gathered} 16) x^{2}y=4(2-y) , (2 , 1) \\ Solution: \\ 2xy+x^2\frac{dy}{dx}=-4\frac{dy}{dx} \\ when x=2 , y=1 \\ 2(2\left)\right(1)+{\left(2\right)}^2\frac{dy}{dx}=-4\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=-\frac{1}{2} \end{gathered}$$            

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             $$\begin{gathered} 17) e^{sinx}+e^{cosy}=e+1 , (\frac{\pi }{2} , \frac{\pi }{2}) \\ Solution: \\ cosx e^{sinx}-siny\frac{dy}{dx} e^{cosy}=0 \\ when x=\frac{\pi }{2} , y=\frac{\pi }{2} \\ cos(\frac{\pi }{2}) e^{sin\left(\frac{\pi }{2}\right)}-sin(\frac{\pi }{2})\frac{dy}{dx} e^{cos\left(\frac{\pi }{2}\right)}=0 \\ \frac{dy}{dx}=0 \end{gathered}$$ 

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             $$\begin{gathered} 18) \sqrt[3]{x^2}+\sqrt[3]{y^2}=5 , (8 , 1) \\ Solution: \\ x^{\frac{2}{3}}+ y^{\frac{2}{3}}=5 \\ \frac{2}{3} x^{\frac{-1}{3}}+\frac{2}{3} y^{\frac{-1}{3}} \frac{dy}{dx}=0 \\ when x=8 , y=1 \\ {\left(8\right)}^{\frac{-1}{3}}+ {\left(1\right)}^{\frac{-1}{3}} \frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{1}{2} \end{gathered}$$   

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أجد معادلة المماس لمنحنى كل علاقة مما يأتي عند النقطة المعطاة:

         $$\begin{gathered} 19) x^2+xy+y^2=13 , (-4 , 3) \\ Solution: \\ 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0 \\ when x= -4 , y=3 \\ 2(-4)+(3)+(-4)\frac{dy}{dx}+2(3)\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}= \frac{5}{2} \\ \Rightarrow y-y_1=\frac{-1}{m}(x-x_1) \\ \Rightarrow y-3=- \frac{2}{5}(x+4) \end{gathered}$$            

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           $$\begin{gathered} 20) x+y-1=ln(x^2+y^2) , (1 , 0) \\ Solution: \\ 1+\frac{dy}{dx}=\frac{2x+2y\frac{dy}{dx}}{x^2+y^2} \\ when x=1 , y=0 \\ 1+\frac{dy}{dx}=\frac{2\left(1\right)+2\left(0\right)\frac{dy}{dx}}{{\left(1\right)}^2+{\left(0\right)}^2} \\ 1+\frac{dy}{dx}=2 \Rightarrow \frac{dy}{dx}=1 \\ \Rightarrow y-y_1=\frac{-1}{m}(x-x_1) \\ \Rightarrow y-0=-(x-1) \Rightarrow y=1-x \end{gathered}$$           

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أجد  $\frac{d^{2}y}{d{x}^2}$ لكلٌّ ممّا يأتي 

           $$\begin{gathered} 21) x+y=siny \\ Solution: \\ 1+\frac{dy}{dx}=cosy\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{cosy-1} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{siny\frac{dy}{dx}}{{(cosy-1)}^2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{siny\left(\frac{1}{cosy-1}\right)}{{(cosy-1)}^2}=\frac{siny}{{(cosy-1)}^3} \end{gathered}$$          

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           $$\begin{gathered} 22) 4y^3=6x^2+1 \\ Solution: \\ 12y^2\frac{dy}{dx}=12x \\ \Rightarrow \frac{dy}{dx}=\frac{x}{y^2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{y^2-2y\frac{dy}{dx}x}{y^4} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{y^2-2yx\left(\frac{x}{y^2}\right)}{y^4}=\frac{y^2-2x\left(\frac{1}{y}\right)}{y^4} \\ \frac{d^{2}y}{d{x}^2}=\frac{y^3-2x}{y^5} \end{gathered}$$          

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        $$\begin{gathered} 23) xy+e^y=e \\ Solution: \\ y+x\frac{dy}{dx}+\frac{dy}{dx}{e}^y=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{y}{ x+e^y} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{\frac{dy}{dx}( x+e^y)-y(1+\frac{dy}{dx}{e}^y)}{ {(x+e^y)}^2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{(-\frac{y}{ x+e^y})( x+e^y)-y(1+(-\frac{y}{ x+e^y})e^y)}{ {(x+e^y)}^2} \\ \frac{d^{2}y}{d{x}^2}=-\frac{-y-y{e}^y+\frac{y^2{e}^y}{ x+e^y}}{ {(x+e^y)}^2} \end{gathered}$$            

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24)  ‏أجد معادلة العمودي على المماس لمنحنى العلاقة: $(x-6)(y+4)=2$عند النقطة‎  $(7 , -2)$

            $$\begin{gathered} Solution: \\ (y+4)+(x-6)\frac{dy}{dx}=0 \\ when x=7 , y=-2 \\ (-2+4)+(7-6)\frac{dy}{dx}=0 \\ \frac{dy}{dx}=-2 \\ \Rightarrow y-y_1=\frac{-1}{m}(x-x_1) \Rightarrow y+2=\frac{1}{2}(x-7) \end{gathered}$$           

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‏25)   أثبت أن لمنحنى العلاقة: $3x^2+2xy+y^2=6$مماسين أفقيين، ثم أجد إحداثيي نقطتي التماس.‎  

                           $$\begin{gathered} Solution: \\ 6x+2y+2x\frac{dy}{dx}+2y\frac{dy}{dx}=0 when \frac{dy}{dx}=0 \\ 6x+2y+2x\left(0\right)+2y\left(0\right)=0 \\ 3x+y=0 \Rightarrow y=-3x \\ 3x^2+2xy+y^2=6 when y=-3x \\ 3x^2+2x(-3x)+{(-3x)}^2=6 \\ 3x^2-6x^2+{9x}^2=6 \\ 6x^2=6 \Rightarrow x= \pm 1 \\ y=-3 x=1 \Rightarrow (1 , -3) \\ y=3 x=-1 \Rightarrow (-1 , 3) \end{gathered}$$

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‏26)  أجد إحداثيي نقطة على المنحنى:  $x+y^2=1$ بحيث يكون عندها مماس المنحنى موازيًا للمستقيم: $x+2y=0$

        $$\begin{gathered} Solution: \\ x+y^2=1 \\ 1+2y\frac{dy}{dx}=0 \\ \Rightarrow 1+2y\frac{dy}{dx}=-\frac{1}{2y} \\ \Rightarrow \frac{dy}{dx}=-\frac{1}{2y} ....... \overline{)1} \\ x+2y=0 \\ 1+2\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{1}{2} ....... \overline{)2} \\ From 1 and 2 \\ \frac{1}{2y}=\frac{1}{2} \Rightarrow y=1 \\ When y=1 \\ x+{\left(1\right)}^2=1 \Rightarrow x=0 the point=(0 , 1) \end{gathered}$$            

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‏27)  أجد إحداثيي نقطة (نقاط) على المنحنى:  $y^3=x^2$،  بحيث يكون عندها مماس المنحنى عموديًا على المستقيم :‎       ، $y\ne 0$ حيث  $y+3x-5=0$

               $$\begin{gathered} Solution: \\ {y}^3=x^2 \\ 3y^2\frac{dy}{dx}=2x \\ \Rightarrow \frac{dy}{dx}=\frac{2x}{3y^2} ....... \overline{)1} \\ y+3x-5=0 \\ \frac{dy}{dx} +3 = 0 \\ \Rightarrow \frac{dy}{dx}=-3 ....... \overline{)2} \\ From 1 \& 2 \\ \frac{2x}{3y^2}\times -3 = -1 \\ \Rightarrow 2x= y^2 ....... \overline{)3} and y^3=x^2 \\ \Rightarrow y^3={\left(\frac{y^2}{2}\right)}^2 \\ \Rightarrow 4y^3-y^4=0 \\ y^3(4-y)=0 \\ \Rightarrow y=0 \Rightarrow x=0 the point=(0 , 0) \\ \Rightarrow y=4 \Rightarrow x=8 the point= (8 , 4) \end{gathered}$$          

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28)  إذا كان:$\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=10$ ، حيث $x> 0 , y> 0$ ، فأثبت أنَّ $\frac{dy}{dx}=\frac{y}{x}$  . 

                $$\begin{gathered} Solution: \\ \sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=10 \\ \frac{x}{y}+\frac{y}{x}+2=100 \\ \frac{x}{y}+\frac{y}{x}=98 \\ \frac{y-x\frac{dy}{dx}}{y^2}+\frac{\frac{dy}{dx}x-y}{x^2}=0 with x^2{y}^2 \\ x^2(y-x\frac{dy}{dx})+y^2(\frac{dy}{dx}x-y)=0 \\ x^{2}y-x^3\frac{dy}{dx}+y^{2}x\frac{dy}{dx}-y^3=0 \\ \frac{dy}{dx}(-x^3+y^{2}x)=-x^{2}y +y^3 \\ \frac{dy}{dx}=\frac{-x^{2}y +y^3}{-x^3+y^{2}x}=\frac{y\cancel{(-x^2+y^2)}}{x\cancel{(-x^2+y^2)}} \\ \frac{dy}{dx}=\frac{y}{x} \end{gathered}$$          

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29)  أجد إحداثيي النقطة على منحنى الاقتران:  $y=x^{\frac{1}{x}} , x>0$  التي يكون عندها ميل المماس صفرًا.

          $$\begin{gathered} Solution: \\ y=x^{\frac{1}{x}} \\ lny=ln{x}^{\frac{1}{x}} \Rightarrow lny=\frac{1}{x}lnx \\ \frac{{\displaystyle \frac{dy}{dx}}}{y}=-\frac{1}{x^2}lnx + \frac{1}{x}\times \frac{1}{x} \\ \frac{dy}{dx}=\frac{y}{x^2}(1-lnx) \\ When \frac{dy}{dx}=0 \\ 0=\frac{y}{x^2}(1-lnx) \\ \Rightarrow 0=y(1-lnx) \\ \Rightarrow y=0 Rejected \\ \Rightarrow 0=1-lnx \Rightarrow 1=lnx \Rightarrow \\ x=e , y=e^{\frac{1}{e}} \end{gathered}$$              

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30)   أجد إحداثيات جميع النقاط على منحنى الدائرة: $x^2+y^2=100$ التي يكون عندها ميل المماس $\frac{3}{4}$ .

          $$\begin{gathered} Solution: \\ x^2+y^2=100 .... \overline{)1} \\ 2x+2y\frac{dy}{dx}=0 When \frac{dy}{dx}=\frac{3}{4} \\ 2x+2y\left(\frac{3}{4}\right)=0 \\ y=-\frac{4}{3}x .... \overline{)2} \\ From \overline{)1} , \overline{)2} : \\ x^2+{(-\frac{4}{3}x)}^2=100 \frac{25}{9}{x}^2=100 \\ \Rightarrow x=6 ,\Rightarrow y=-\frac{4}{3}\left(6\right) =-8 \Rightarrow (6 ,-8) \\ \Rightarrow x=-6 ,\Rightarrow y=-\frac{4}{3}(-6) =8 \Rightarrow (-6 ,8) \end{gathered}$$              

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31)  يمُثل الاقتران: $s\left(t\right)=t^{\frac{1}{t}} , t>0$ موقع جُسَيْم يتحرّك في مسار مستقيم؛ حيث s الموقع بالأمتار، وt  الزمن بالثواني.

        أجد سرعة الجُسَيْم وتسارعه. 

                  $$\begin{gathered} Solution: \\ s\left(t\right)=t^{\frac{1}{t}} , t>0 \\ ln\left(s\right(t\left)\right)=ln\left(t^{\frac{1}{t}}\right) \\ ln\left(s\right(t\left)\right)=\frac{1}{t}ln\left(t\right) \\ \frac{s^{\text{'}}\left(t\right)}{s\left(t\right)}=-\frac{1}{t^2}ln\left(t\right) + \frac{1}{t}\times \frac{1}{t} \\ s^{\text{'}}\left(t\right)=\frac{s\left(t\right)}{t^2}(1-ln(t\left)\right) \\ v\left(t\right)=t^{(\frac{1}{t}-2)}(1-ln(t\left)\right) \\ a\left(t\right)=v^{\text{'}}\left(t\right)=\left((\frac{1}{t}-2)t^{(\frac{1}{t}-3)}+ (-\frac{1}{t^2})t^{(\frac{1}{t}-2)}\times lnt\right)(1-ln(t\left)\right)+\left(t^{(\frac{1}{t}-2)}(-\frac{1}{t})\right) \end{gathered}$$          

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32)  إذا كان $y=lnx$، حيث $x > 0$ ، فأثبت أنَّ $\frac{dy}{dx}=\frac{1}{x}$باستخدام الاشتقاق الضمني .

$$\begin{gathered} Solution: \\ y=ln x \\ \frac{dy}{dx}=\frac{1}{x} \end{gathered}$$              

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أجد مشتقة مشتقة كل من الاقترانات الآتية باستعمال الاشتقاق اللوغاريتمي

           $$\begin{gathered} 33) y={(x^2+3)}^x \\ Solution: \\ lny=ln{(x^2+3)}^x \\ lny=xln(x^2+3) \\ \frac{{\displaystyle \frac{dy}{dx}}}{y}=ln(x^2+3) +\frac{2x}{x^2+3} \\ \frac{dy}{dx}=y(ln(x^2+3) +\frac{2x}{x^2+3}) \\ \frac{dy}{dx}=2x{(x^2+3)}^{x-1} +{(x^2+3)}^x(ln(x^2+3) \end{gathered}$$            

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             $$\begin{gathered} 34) y=\frac{(x^4+1)\sqrt{x+2}}{2x^2+2x+1} \\ Solution: \\ ln y=ln\left(\frac{(x^4+1)\sqrt{x+2}}{2x^2+2x+1}\right) \\ ln y=ln\left(\right(x^4+1\left)\sqrt{x+2}\right)-ln(2x^2+2x+1) \\ ln y=ln\left(\right(x^4+1)+ln{(x+2)}^{\frac{1}{2}}-ln(2x^2+2x+1) \\ lny=ln\left(\right(x^4+1)+\frac{1}{2}ln(x+2)-ln(2x^2+2x+1) \\ \frac{\frac{dy}{dx}}{y}=\frac{4x^3}{x^4+1}+\frac{1}{2(x+2)}-\frac{4x+2}{2x^2+2x+1} \\ \frac{dy}{dx}=\left(\frac{(x^4+1)\sqrt{x+2}}{2x^2+2x+1}\right)(\frac{4x^3}{x^4+1}+\frac{1}{2(x+2)}-\frac{4x+2}{2x^2+2x+1}) \end{gathered}$$          

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             $$\begin{gathered} \mathit{35}\mathit{)}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }y\mathit{=}\sqrt{x^{\mathit{2}}\mathit{(}x\mathit{+}\mathit{1}\mathit{)}\mathit{(}x\mathit{+}\mathit{2}\mathit{)}} \\ Solution\mathit{:} \\ \mathit{ln}y\mathit{=}\mathit{ln}\mathit{(}{x}^{\mathit{2}}\mathit{(}x\mathit{+}\mathit{1}\mathit{)}{\mathit{(}x\mathit{+}\mathit{2}\mathit{)}\mathit{)}}^{\frac{\mathit{1}}{\mathit{2}}}\mathit{ }lny\mathit{=}\frac{\mathit{1}}{\mathit{2}}ln\mathit{\left(}{x}^{\mathit{2}}\mathit{\right(}x\mathit{+}\mathit{1}\mathit{)}\mathit{(}x\mathit{+}\mathit{2}\mathit{)}\mathit{)} \\ lny\mathit{=}\frac{\mathit{1}}{\mathit{2}}ln\mathit{(}{x}^{\mathit{2}}\mathit{)}\mathit{+}\mathit{ln}\mathit{(}x\mathit{+}\mathit{1}\mathit{)}\mathit{+}\mathit{ln}\mathit{(}x\mathit{+}\mathit{2}\mathit{)} \\ lny\mathit{=}ln\mathit{(}x\mathit{)}\mathit{+}ln\mathit{(}x\mathit{+}\mathit{1}\mathit{)}\mathit{+}ln\mathit{(}x\mathit{+}\mathit{2}\mathit{)} \\ \frac{{\displaystyle \frac{dy}{dx}}}{y}\mathit{=}\frac{\mathit{1}}{x}\mathit{+}\frac{\mathit{1}}{x\mathit{+}\mathit{1}}\mathit{+}\frac{\mathit{1}}{x\mathit{+}\mathit{2}} \\ \frac{dy}{dx}\mathit{=}\sqrt{x^{\mathit{2}}\mathit{(}x\mathit{+}\mathit{1}\mathit{)}\mathit{(}x\mathit{+}\mathit{2}\mathit{)}}\mathit{(}\frac{\mathit{1}}{x}\mathit{+}\frac{\mathit{1}}{x\mathit{+}\mathit{1}}\mathit{+}\frac{\mathit{1}}{x\mathit{+}\mathit{2}}\mathit{)} \end{gathered}$$         

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           $$\begin{gathered} 36) y=x^{sinx} x>0 \\ Solution: \\ lny=ln{x}^{sinx} \\ lny=sinx lnx \\ \frac{\frac{dy}{dx}}{y}=cosx lnx +sinx \times \frac{1}{x} \\ \frac{dy}{dx}=x^{sinx}(cosx lnx +\frac{sinx }{x}) \end{gathered}$$ 

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أجد  $\frac{d^{2}y}{d{x}^2}$ لكل معادلة وسيطية مما يأتي عند قيمة  $x$ المعطاة:

            $$\begin{gathered} 37) x=sint , y=cost , t=\frac{\pi }{4} \\ Solution: \\ x=sint \Rightarrow \frac{dx}{dt}=cost \\ y=cost \Rightarrow \frac{dy}{dt}=-sint \\ \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx} \\ =-sint \times \frac{1}{cost}=-tant \\ \frac{d}{dt}\left(\frac{dy}{dx}\right)=-se{c}^{2}t \\ \frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{-se{c}^{2}t }{cost} \\ When t=\frac{\pi }{4} \\ \frac{d^{2}y}{d{x}^2}=\frac{-se{c}^2\left(\frac{\pi }{4}\right) }{cos\left(\frac{\pi }{4}\right)}=-2\sqrt{2} \end{gathered}$$

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            $$\begin{gathered} \mathit{ }\mathit{38}\mathit{)}\mathit{ }\mathit{ }\mathit{ }\mathit{ }x\mathit{=}{e}^{\mathit{-}t}\mathit{ }\mathit{,}\mathit{ }y\mathit{=}{t}^{\mathit{3}}\mathit{+}t\mathit{+}\mathit{1}\mathit{ }\mathit{ }\mathit{,}t\mathit{=}\mathit{0}\mathit{ }\mathit{ } \\ Solution\mathit{:} \\ \mathit{ }x\mathit{=}{e}^{\mathit{-}t}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }\frac{dx}{dt}\mathit{=}\mathit{-}{e}^{\mathit{-}t}\mathit{ } \\ \mathit{ }y\mathit{=}{t}^{\mathit{3}}\mathit{+}t\mathit{+}\mathit{1}\mathit{ }\mathit{ }\mathit{ }\mathit{\Rightarrow }\frac{dy}{dt}\mathit{=}\mathit{3}{t}^{\mathit{2}}\mathit{+}\mathit{1}\mathit{ } \\ \frac{dy}{dx}\mathit{=}\frac{dy}{dt}\mathit{\times }\frac{dt}{dx} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{(}\mathit{3}{t}^{\mathit{2}}\mathit{+}\mathit{1}\mathit{)}\mathit{ }\mathit{\times }\frac{\mathit{1}}{\mathit{-}{e}^{\mathit{-}t}}\mathit{=}\mathit{-}{e}^t\mathit{(}\mathit{3}{t}^{\mathit{2}}\mathit{+}\mathit{1}\mathit{)}\mathit{ } \\ \frac{d}{dt}\mathit{\left(}\frac{dy}{dx}\mathit{\right)}\mathit{=}\frac{d}{dt}\mathit{(}\mathit{-}{e}^t\mathit{(}\mathit{3}{t}^{\mathit{2}}\mathit{+}\mathit{1}\mathit{\left)}\mathit{\right)}\mathit{ } \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{=}\mathit{-}{e}^t\mathit{(}\mathit{3}{t}^{\mathit{2}}\mathit{+}\mathit{1}\mathit{)}\mathit{+}\mathit{-}{e}^t\mathit{(}\mathit{6}t\mathit{)} \\ \frac{d^{\mathit{2}}y}{d{x}^{\mathit{2}}}\mathit{=}\frac{\frac{d}{dt}\mathit{\left(}\frac{dy}{dx}\mathit{\right)}}{\frac{dx}{dt}}\mathit{=}\frac{\mathit{-}{e}^t\mathit{(}\mathit{3}{t}^{\mathit{2}}\mathit{+}\mathit{1}\mathit{)}\mathit{+}\mathit{-}{e}^t\mathit{\left(}\mathit{6}t\mathit{\right)}\mathit{ }}{\mathit{-}{e}^{\mathit{-}t}} \\ When\mathit{ }\mathit{ }t\mathit{=}\mathit{0} \\ \frac{d^{\mathit{2}}y}{d{x}^{\mathit{2}}}\mathit{=}\frac{\mathit{-}{e}^{\mathit{\left(}\mathit{0}\mathit{\right)}}\mathit{(}\mathit{3}{\mathit{\left(}\mathit{0}\mathit{\right)}}^{\mathit{2}}\mathit{+}\mathit{1}\mathit{)}\mathit{+}\mathit{-}{e}^{\mathit{\left(}\mathit{0}\mathit{\right)}}\mathit{\left(}\mathit{6}\mathit{\right(}\mathit{0}\mathit{\left)}\mathit{\right)}\mathit{ }}{\mathit{-}{e}^{\mathit{-}\mathit{\left(}\mathit{0}\mathit{\right)}}}\mathit{=}\frac{\mathit{-}{e}^{\mathit{\left(}\mathit{0}\mathit{\right)}}\mathit{ }}{\mathit{-}{e}^{\mathit{-}\mathit{\left(}\mathit{0}\mathit{\right)}}}\mathit{=}\mathit{1} \end{gathered}$$

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إذا كانت العلاقة: $x^3+y^3=6xy$ فأجيب عن السؤالين الآتيين تباعًا

39)  أجد معادلة المماس عند نقطة تقاطع منحنى المعادلة مع منحنى: $y=x$  في الربع الأوّل.

         $$\begin{gathered} Solution: \\ x^3+y^3=6xy ..... \overline{)1} \\ y=x ..... \overline{)2} \\ x^3+{\left(x\right)}^3=6x\left(x\right) \\ 2x^3- 6x^2=0 \\ 2x^2(x- 3)=0 \\ \Rightarrow x=0 \Rightarrow y=0 \\ \Rightarrow x=3 \Rightarrow y=3 \\ m=\frac{dy}{dx} When x=3 , y=3 \\ 3x^2+3y^2\frac{dy}{dx}=6y +6x\frac{dy}{dx} \\ {\left(3\right)}^2+{\left(3\right)}^2\frac{dy}{dx}=2\left(3\right) +2\left(3\right)\frac{dy}{dx} \\ \frac{dy}{dx}=-1 \\ y-y_1=m(x-x_1) \\ y-3=-1(x-3) \Rightarrow y=6-x \end{gathered}$$           

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40)  أجد إحداثيي نقطة على منحنى العلاقة $x^3+y^3=6xy$في الربع الأوّل؛ بحيث يكون عندها مماس المنحنى أفقيًا.

            $$\begin{gathered} Solution: \\ x^3+y^3=6xy , \frac{dy}{dx}=0 .... \overline{)1} \\ x^2+y^2\frac{dy}{dx}=2y +2x\frac{dy}{dx} \\ x^2+y^2\left(0\right)=2y +2x\left(0\right) \\ \Rightarrow x^2=2y \Rightarrow y=\frac{x^2}{2} .... \overline{)2} \\ x^3+{\left(\frac{x^2}{2}\right)}^3=6x\left(\frac{x^2}{2}\right) \\ x^3+\frac{x^6}{8}=3x^3 \\ {x}^6-16x^3=0 \\ x^3( x^3-16)=0 \\ \Rightarrow x=0 \Rightarrow y=0 \Rightarrow (0 , 0) \\ \Rightarrow x=\sqrt[3]{16} \Rightarrow y=\frac{{\left(\sqrt[3]{16}\right)}^2}{2} \Rightarrow (\sqrt[3]{16} ,\frac{{\left(\sqrt[3]{16}\right)}^2}{2} ) \end{gathered}$$             

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سنجد معادلة المماس من بمعلومية نقطة التماس وهي $(x ,\sqrt{2-x^2})$

$$\begin{gathered} \mathit{ }Solution\mathit{:} \\ {x}^{\mathit{2}}+y^{\mathit{2}}\mathit{=}\mathit{2}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{ }\overline{)\mathit{1}} \\ \mathit{2}x\mathit{ }\mathit{+}\mathit{2}y\mathit{ }\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\mathit{=}\mathit{0} \\ but\mathit{ }\frac{dy}{dx}\mathit{=}\mathit{tan}\mathit{\left(}\frac{\mathit{3}\mathit{\pi }}{\mathit{4}}\mathit{\right)}\mathit{=}\mathit{-}\mathit{1} \\ \mathit{\Rightarrow }\mathit{2}x\mathit{ }\mathit{+}\mathit{2}y\mathit{(}\mathit{-}\mathit{1}\mathit{)}\mathit{=}\mathit{0} \\ \mathit{\Rightarrow }\mathit{ }y\mathit{=}x\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{ }\overline{)\mathit{2}} \\ From\mathit{ }\mathit{ }\overline{)\mathit{1}}\mathit{ }\mathit{ }and\mathit{ }\mathit{ }\overline{)\mathit{2}} \\ y^2+y^2=2\mathit{ }\mathit{ }\mathit{\Rightarrow }\mathit{ }{y}^2=1\mathit{ } \\ when\mathit{ }\mathit{ }y\mathit{=}\mathit{1}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }then\mathit{ }x\mathit{=}\mathit{1}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }then\mathit{ }\mathit{ }\mathit{(}\mathit{1}\mathit{ }\mathit{,}\mathit{ }\mathit{1}\mathit{)} \\ \mathit{ }y\mathit{-}{y}_{\mathit{1}}\mathit{=}m\mathit{(}x\mathit{-}{x}_{\mathit{1}}\mathit{)} \\ \mathit{ }y\mathit{-}\mathit{1}\mathit{=}\mathit{-}\mathit{1}\mathit{(}x\mathit{-}\mathit{1}\mathit{)} \\ when\mathit{ }\mathit{ }y\mathit{=}\mathit{-}\mathit{1}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }then\mathit{ }x\mathit{=}\mathit{-}\mathit{1}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }then\mathit{ }\mathit{ }\mathit{(}\mathit{-}\mathit{1}\mathit{ }\mathit{,}\mathit{-}\mathit{ }\mathit{1}\mathit{)} \\ y\mathit{-}{y}_{\mathit{1}}\mathit{=}m\mathit{(}x\mathit{-}{x}_{\mathit{1}}\mathit{)} \\ y\mathit{+}\mathit{1}\mathit{=}\mathit{-}\mathit{1}\mathit{(}x\mathit{+}\mathit{1}\mathit{)} \end{gathered}$$            

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تبرير: إذا كان: $\mathit{ }{x}^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}\mathit{=}\mathit{1}$،  فأجيب عن الأسئلة الأربعة الآتية تباعًا:

42)  أجد $\frac{dy}{dx}$

           $$\begin{gathered} Solution: \\ {x}^2-y^2=1 \\ 2x-2y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{x}{y} \end{gathered}$$   

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43)  يُمكن التعبير عن منحنى العلاقة:  $\mathit{ }{x}^{\mathit{2}}\mathit{-}{y}^{\mathit{2}}\mathit{=}\mathit{1}$بالمعادلة الوسيطية:

        $x=sect , y=tant$. حيث : $-\frac{\mathrm{\pi }}{2}\le t\le \frac{\mathrm{\pi }}{2}$ . أستعمل هذه الحقيقة لإيجاد $\frac{dy}{dx}$  بدلالة t .

          $$\begin{gathered} Solution: \\ x=sect \Rightarrow \frac{dx}{dt}=sect tant \\ y=tant \Rightarrow \frac{dy}{dt}=se{c}^{2}t \\ \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx} \\ =(se{c}^{2}t ) \times \frac{1}{sect tant}=\frac{sect}{ tant} \end{gathered}$$           

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44)   أثبت أن المقدارين : الجبريين اللذين يُمثّلان $\frac{dy}{dx}$ الناتجين  في الفرعين السابقين متكافئان ، مُبرَّرًا إجابتي.

                $$\begin{gathered} \mathit{ }Solution\mathit{:} \\ \frac{dy}{dx}\mathit{=}\frac{x}{y}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }from\mathit{ }\overline{)\mathit{42}} \\ \frac{dy}{dx}\mathit{=}\frac{sect}{tant}\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }from\mathit{ }\overline{)\mathit{43}} \\ but\mathit{ }\mathit{ }\mathit{ }x\mathit{=}\mathit{ }sect \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }y\mathit{=}\mathit{ }tant\mathit{ } \\ So\mathit{ }\overline{)\mathit{42}}\mathit{=}\overline{)\mathit{43}} \end{gathered}$$          

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45)  أجد إحداثيات النقاط التي يكون عندها ميل المماس 2.

           $$\begin{gathered} Solution: \\ \frac{dy}{dx}=\frac{sect}{ tant}=2 \\ \Rightarrow \frac{sect}{ tant}=\frac{{\displaystyle \frac{1}{cost}}}{{\displaystyle \frac{sint}{cost}}}=2 \\ \Rightarrow \frac{1}{sint}=2 \\ \Rightarrow \frac{1}{2}=sint \Rightarrow t=\frac{\pi }{6} , \frac{5\pi }{6} \\ when t=\frac{\pi }{6} \Rightarrow x= sec\frac{\pi }{6} = \frac{2}{\sqrt{3}} , and y=tan\frac{\pi }{6}=\frac{1}{\sqrt{3}} \Rightarrow ( \frac{2}{\sqrt{3}} , \frac{1}{\sqrt{3}}) \\ when t=\frac{5\pi }{6} \Rightarrow x= sec\frac{5\pi }{6} = -\frac{2}{\sqrt{3}} , and y=tan\frac{5\pi }{6}=-\frac{1}{\sqrt{3}} \Rightarrow ( -\frac{2}{\sqrt{3}} , -\frac{1}{\sqrt{3}}) \end{gathered}$$           

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46)  تبرير: إذا مثل  l  أيَّ مماس لمنحنى المعادلة: $\sqrt{x}+\sqrt{y}=\sqrt{k}$ حيث k  ثابت موجب،  

       فأثبت أن مجموع المقطع   x  والمقطع y للمستقيم  l  يساوي  k مُبرَّرًا إجابتي.

       سنجد معادلة المماس l عند النقطة (a ,b)

        $$\begin{gathered} Solution: \\ \sqrt{x}+\sqrt{y}=\sqrt{k} \\ \frac{1}{2\sqrt{x}}+\frac{\frac{dy}{dx}}{2\sqrt{y}}=0 \\ \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}} \\ Now m=\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}} at (a , b) \\ y-y_1=m(x-x_1) \\ y-b=-\frac{\sqrt{y}}{\sqrt{x}}(x-a) \\ now when b=0 \\ y=-\frac{\sqrt{y}}{\sqrt{x}}(x-a) \\ a=x+\sqrt{xy} .... \overline{)1} \\ now when a=0 \\ y-b=-\frac{\sqrt{y}}{\sqrt{x}}\left(x\right) \\ b=y+\sqrt{xy} .... \overline{)2} \\ \overline{)1} + \overline{)2} =a+b \\ a+b = x+\sqrt{xy} + y+\sqrt{xy} \\ = x+2\sqrt{xy} + y =k \\ \Rightarrow a+b=k \end{gathered}$$           

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47)  تحدّ: إذا كان مماس منحنى الاقتران: $y={(x-3)}^{\sqrt{x}}$عند النقطة $(4 , 1)$ يقطع المحور x في النقطة  B  ،

            والمحور y   في النقطة  C ؛ فأجد مساحة $∆OBC$ حيث O نقطة الأصل.

            سنجد معادلة المماس  عند النقطة  $(4 , 1)$

                        $$\begin{gathered} Solution: \\ y={(x-3)}^{\sqrt{x}} \\ lny=ln {(x-3)}^{\sqrt{x}} \\ lny=\sqrt{x} ln( x-3) \\ \frac{\frac{dy}{dx}}{y}=\frac{ln (x-3)}{2\sqrt{x}} + \frac{\sqrt{x}}{x-3} \\ when x=4 , y=1 \\ \frac{\frac{dy}{dx}}{1}=\frac{ln (4-3)}{2\sqrt{4}} + \frac{\sqrt{4}}{4-3} \\ \frac{dy}{dx}=0+2=2 \\ Now m=\frac{dy}{dx}=2 \\ y-1=2(x-4) \\ now when y=0 \\ 0-1=2(x-4) \Rightarrow x=\frac{7}{2} \\ now when x=0 \\ y-1=2(0-4) \Rightarrow y=-7 \\ now \\ A=\frac{1}{2}\left(\frac{7}{2}\right)(\left|-7\right| )=\frac{49}{4} \\ =2(\frac{-1}{2+ln 4}+1)(4+4ln 4 ) \end{gathered}$$