Question

$y = { \sin}^{ - 1} ( {x}^{3} )$
diff wrt x​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given:

$\rm \longrightarrow y = \sin^{ - 1} ( {x}^{3} )$

$\rm \longrightarrow y^{ \prime} = \dfrac{d}{dx}\sin^{ - 1} ( p ) \times \dfrac{d}{dx} (p ) \: \: \: \: \bigg(p = {x}^{3} \bigg)$

$\rm \longrightarrow y^{ \prime} = \dfrac{1}{ \sqrt{1 - {p}^{2} } } \times \dfrac{d}{dx} ( {x}^{3} )$

$\rm \longrightarrow y^{ \prime} = \dfrac{1}{ \sqrt{1 - {p}^{2} } } \times 3 {x}^{2}$

$\rm \longrightarrow y^{ \prime} = \dfrac{1}{ \sqrt{1 - {x}^{6} } } \times 3 {x}^{2}$

$\rm \longrightarrow y^{ \prime} = \dfrac{3 {x}^{2} }{ \sqrt{1 - {x}^{6} } }$

★ Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

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