Question

y=(x^2+4)^5 find dy/dx​

Answer 1

Solution:

Given That:

$\rm \longrightarrow y = {( {x}^{2} + 4)}^{5}$

Taking derivative on both sides, we get:

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx} {( {x}^{2} + 4)}^{5}$

$\rm \longrightarrow \dfrac{dy}{dx} = 5 {( {x}^{2} + 4 )}^{5 - 1} \times \dfrac{d}{dx} ( {x}^{2} + 4)$

$\rm \longrightarrow \dfrac{dy}{dx} = 5 {( {x}^{2} + 4 )}^{4} \times \dfrac{d}{dx} ( {x}^{2} + 4)$

$\rm \longrightarrow \dfrac{dy}{dx} = 5 {( {x}^{2} + 4 )}^{4} \times 2x$

$\rm \longrightarrow \dfrac{dy}{dx} = 10x{( {x}^{2} + 4 )}^{4}$

★ Which is our required answer.

Answer:

$\rm \hookrightarrow \dfrac{dy}{dx} = 10x{( {x}^{2} + 4 )}^{4}$

Learn More:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$