إجابات كتاب التمارين

قاعدة السلسلة

أجد مشتقة كل اقتران ممّا يأتي:

(1) $f\left(x\right)=\sqrt{4x-1}$

$f^{\mathrm{\prime }}\left(x\right)=\frac{4}{2\sqrt{4x-1}}=\frac{2}{\sqrt{4x-1}}$

(2) $f\left(x\right)=\frac{3}{\sqrt{3-x^2}}$

$\begin{array}{rl}& f\left(x\right)=3{(3-x^2)}^{-\frac{1}{2}}\\ & f^{\mathrm{\prime }}\left(x\right)=-\frac{3}{2}{(3-x^2)}^{-\frac{3}{2}}(-2x)=\frac{3x}{\sqrt{{(3-x^2)}^3}}\end{array}$

(3) $f\left(x\right)=(3+4x{)}^{\frac{5}{2}}$

$f^{\mathrm{\prime }}\left(x\right)=\frac{5}{2}(3+4x{)}^{\frac{3}{2}}(4)=10(3+4x{)}^{\frac{3}{2}}$

(4) $f\left(x\right)=(8-x{)}^{100}$

$f^{\mathrm{\prime }}\left(x\right)=100(8-x{)}^{99}(-1)=-100(8-x{)}^{99}$

(5) $f\left(x\right)=x^2+(200-x{)}^2$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(x\right)=2x+2(200-x{)}^1(-1)& =2x-2(200-x)\\ & =2x-400+2x\\ & =4x-400\end{array}$

(6) $f\left(x\right)=(x+5{)}^7+(2x+3{)}^6$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(x\right)& =7(x+5{)}^6(1)+6(2x+3{)}^5\left(2\right)\\ & =7(x+5{)}^6+12(2x+3{)}^5\end{array}$

(7) $f\left(x\right)=\sqrt[3]{x^5+6x}$

$\begin{array}{rl}& f\left(x\right)={(x^5+6x)}^{\frac{1}{3}}\\ & f^{\mathrm{\prime }}\left(x\right)=\frac{1}{3}{(x^5+6x)}^{-\frac{2}{3}}(5x^4+6)=\frac{5x^4+6}{3\sqrt[3]{{(5x^4+6x)}^2}}\end{array}$

(8) $f\left(x\right)=\frac{1}{{(x^2-3)}^3}$

$\begin{array}{rl}& f\left(x\right)={(x^2-3)}^{-3}\\ & f^{\mathrm{\prime }}\left(x\right)=-3{(x^2-3)}^{-4}\left(2x\right)=\frac{-6x}{{(x^2-3)}^4}\end{array}$

(9) $f\left(x\right)=\frac{1}{2}{x}^2+\sqrt{16-x^2}$

$f^{\mathrm{\prime }}\left(x\right)=x+\frac{-2x}{2\sqrt{16-x^2}}=x-\frac{x}{\sqrt{16-x^2}}$

أجد مشتقة كل اقتران ممّا يأتي عند قيمة x المعطاة:

(10) $f\left(x\right)=4x^3+(x-2{)}^4 , x=2$

$\begin{array}{rl}& f^{\mathrm{\prime }}\left(x\right)=12x^2+4(x-2{)}^3\\ & f^{\mathrm{\prime }}\left(2\right)=12\left(4\right)+4(2-2{)}^3=48\end{array}$

(11) $f\left(x\right)=\sqrt{x^2+8x }, x=8$

$\begin{array}{rl}& f^{\mathrm{\prime }}\left(x\right)=\frac{2x+8}{2\sqrt{x^2+8x}}=\frac{x+4}{\sqrt{x^2+8x}}\\ & f^{\mathrm{\prime }}\left(8\right)=\frac{8+4}{\sqrt{64+64}}=\frac{12}{\sqrt{128}}=\frac{12}{8\sqrt{2}}=\frac{3}{2\sqrt{2}}\end{array}$

أستعمل قاعدة السلسلة في إيجاد $\frac{dy}{dx}$ لكلّ ممّا يأتي:

(12) $y=u^3-7u^2 , u=x^2+3$

$\begin{array}{rl}\frac{dy}{du}& =3u^2-14u\\ \frac{du}{dx}& =2x\\ \frac{dy}{dx}& =\frac{dy}{du}\times \frac{du}{dx}\\ & =(3u^2-14u)\times 2x\\ & =6x{(x^2+3)}^2-28x(x^2+3)\end{array}$

(13) $y=\sqrt{7-3u }, u=x^2-9$

$\begin{array}{rl}\frac{dy}{du}& =\frac{-3}{2\sqrt{7-3u}}\\ \frac{du}{dx}& =2x\\ \frac{dy}{dx}& =\frac{dy}{du}\times \frac{du}{dx}\\ & =\frac{-3}{2\sqrt{7-3u}}\times 2x\\ & =\frac{-3x}{\sqrt{7-3(x^2-9)}}=\frac{-3x}{\sqrt{34-3x^2}}\end{array}$

أستعمل قاعدة السلسلة في إيجاد $\frac{dy}{dx}$ لكلّ ممّا يأتي عند قيمة x المعطاة:

(14) $f\left(x\right)=u^3-5{(u^3-7u)}^2 , u=\sqrt{x} , x=4$

$\begin{array}{rl}& \frac{dy}{du}=3u^2-10(u^3-7u)(3u^2-7)\\ & \frac{du}{dx}=\frac{1}{2\sqrt{x}}\end{array}$

عندما x = 4 ، فإن u = 2

$\begin{array}{rl}& {\frac{dy}{dx}|}_{x=4}={\frac{dy}{du}|}_{u=2}\times {\frac{du}{dx}|}_{x=4}\\ & {\frac{dy}{du}|}_{u=2}=3\left(4\right)-10(8-14)(12-7)=312\\ & {\frac{du}{dx}|}_{x=4}=\frac{1}{4}\\ & {\frac{dy}{dx}|}_{x=4}=312\times \frac{1}{4}=78\end{array}$

(15) $f\left(x\right)=2u^3+3u^2 , u=x+\sqrt{x }, x=1$

$\begin{array}{rl}& \frac{dy}{du}=6u^2+6u\\ & \frac{du}{dx}=1+\frac{1}{2\sqrt{x}}\end{array}$

عندما x = 1 ، فإن u = 2

$\begin{array}{rl}{\frac{dy}{dx}|}_{x=1}& ={\frac{dy}{du}|}_{u=2}\times {\frac{du}{dx}|}_{x=1}\\ {\frac{dy}{du}|}_{u=2}& =6\left(4\right)+6\left(2\right)=36\\ {\frac{du}{dx}|}_{x=1}& =1+\frac{1}{2}=\frac{3}{2}\\ {\frac{dy}{dx}|}_{x=1}& =36\times \frac{3}{2}=54\end{array}$

تلوث: توصلت دراسة بيئية إلى نمذجة مقدار التلوث في إحدى البحيرات باستعمال الاقتران: $\mathit{P}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathbf{(}{\mathbf{t}}^{\frac{\mathbf{1}}{\mathbf{4}}}\mathbf{+}\mathbf{3}\mathbf{)}}^{\mathbf{3}}$ ، حيث t الزمن بالسنوات، علماً بأنّ P يقاس بأجزاء من المليون:

(16) أجد معدل تغير مقدار التلوث في البحيرة بالنسبة إلى الزمن t .

$P^{\mathrm{\prime }}\left(t\right)=3{(t^{\frac{1}{4}}+3)}^2\times \frac{1}{4}{t}^{-\frac{3}{4}}=\frac{3{(t^{\frac{1}{4}}+3)}^2}{4t^{\frac{3}{4}}}$

(17) أجد معدل تغير مقدار التلوث في البحيرة بعد 16 عاماً.

$P^{\mathrm{\prime }}\left(16\right)=\frac{3(2+3{)}^2}{4\left(8\right)}=\frac{75}{32}\approx 2.34$

إذا كان: $\mathbf{g}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{8}\mathbf{ }\mathbf{,}\mathbf{ }{\mathbf{g}}^{\mathbf{\prime }}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{h}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{ }\mathbf{,}\mathbf{ }{\mathbf{h}}^{\mathbf{\prime }}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{=}\mathbf{6}$ فأجد مشتقة كل اقتران مما يأتي عندما x = 5 :

(18) $f\left(x\right)=g\left(h\right(x\left)\right)$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(x\right)& =g^{\mathrm{\prime }}\left(h\right(x\left)\right)\times h^{\mathrm{\prime }}\left(x\right)\\ f^{\mathrm{\prime }}\left(5\right)& =g^{\mathrm{\prime }}\left(h\right(5\left)\right)\times h^{\mathrm{\prime }}\left(5\right)\\ & =g^{\mathrm{\prime }}(-2)\times 6\\ & =4\times 6=24\end{array}$

(19) $f\left(x\right)=4\left(h\right(x){)}^2$

$\begin{array}{rl}{f}^{\mathrm{\prime }}\left(x\right)& =8\left(h\right(x\left)\right)\times h^{\mathrm{\prime }}\left(x\right)\\ f^{\mathrm{\prime }}\left(5\right)& =8\left(h\right(5\left)\right)\times h^{\mathrm{\prime }}\left(5\right)\\ & =8\times -2\times 6=-96\end{array}$