Question

Find the 1st derivative of Y = Sin²(arctan x) + Cos²(arctan x) + 5x³​

Answer 1

Solution:

To Determine: The derivative of y wrt x.

Given :-

$\rm \longrightarrow y = { \sin}^{2}( \tan^{ - 1} (x) ) + { \cos}^{2}( { \tan}^{ - 1}(x) ) + 5 {x}^{3}$

We know that :-

$\rm \longrightarrow {\sin}^{2}(x) + \cos^{2} (x) = 1$

Therefore :-

$\rm \longrightarrow y = 1+ 5 {x}^{3}$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx}( 1+ 5 {x}^{3})$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx}( 1)+\dfrac{d}{dx} ( 5 {x}^{3})$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d}{dx}( 1)+5\dfrac{d}{dx} ({x}^{3})$

We know that :-

$\rm (1) \: \: \dfrac{d}{dx}(k) = 0$

$\rm (2) \: \: \dfrac{d}{dx}( {x}^{n} ) =n {x}^{n - 1}$

Therefore, we get :-

$\rm \longrightarrow \dfrac{dy}{dx} =0+5 \times 3 {x}^{3 - 1}$

$\rm \longrightarrow \dfrac{dy}{dx} =15{x}^{2}$

★ Which is our required answer.

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