Question

differentiate w.r.t. of x tan (sin^-1 x^4)
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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:tan\bigg( {sin}^{ - 1} {x}^{4} \bigg)$

Let we assume that

$\rm :\longmapsto\:y \: = \: tan\bigg( {sin}^{ - 1} {x}^{4} \bigg)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y \: = \dfrac{d}{dx} \: tan\bigg( {sin}^{ - 1} {x}^{4} \bigg)$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}tanx = {sec}^{2}x}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = {sec}^{2} \bigg( {sin}^{ - 1} {x}^{4} \bigg)\dfrac{d}{dx}\bigg( {sin}^{ - 1} {x}^{4} \bigg)$

We know

$\boxed{ \bf{ \: \dfrac{d}{dx} {sin}^{ - 1}x = \frac{1}{ \sqrt{1 - {x}^{2} } }}}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = {sec}^{2} \bigg( {sin}^{ - 1} {x}^{4} \bigg) \times \dfrac{1}{ \sqrt{1 - {( {x}^{4} )}^{2} } }\dfrac{d}{dx} {x}^{4}$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \dfrac{{sec}^{2} \bigg( {sin}^{ - 1} {x}^{4} \bigg)}{ \sqrt{1 - {x}^{8} } } \times {4x}^{4 - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \dfrac{{sec}^{2} \bigg( {sin}^{ - 1} {x}^{4} \bigg)}{ \sqrt{1 - {x}^{8} } } \times {4x}^{3}$

$\bf :\longmapsto\:\dfrac{dy}{dx} \: = \dfrac{ {4x}^{3} \: {sec}^{2} \bigg( {sin}^{ - 1} {x}^{4} \bigg)}{ \sqrt{1 - {x}^{8} } }$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$