Question

Find the Derivative with respect to x
$\rm y = {a}^{ (\sin {}^{ - 1} x) {}^{2} } \\ \\ \rm \: find \: \: \dfrac{dy}{dx}$
​

Answer 1

ANSWER :–

GIVEN :–

$\\ \implies { \bold{ y = {a}^{ (\sin ^{ - 1} x)^{2} }}}$

TO FIND :–

$\\ \implies { \bold{ \dfrac{dy}{dx} = ? }}$

SOLUTION :–

$\\ \implies { \bold{ y = {a}^{ (\sin ^{ - 1} x)^{2} }}}$

• Let's take log on both sides –

$\\ \implies { \bold{ log(y) = log [ {a}^{ (\sin ^{ - 1} x)^{2} }]}}$

• Using identity –

$\\ \bigstar \: \large { \orange{ \boxed{ \bold{ log( {a}^{x} ) = (x)log (a)}}}}$

• So that –

$\\ \implies { \bold{ log(y) = { (\sin ^{ - 1} x)^{2} } log (a)}}$

• Now Differentiate with respect to 'x' –

$\\ \implies { \bold{ \dfrac{1}{y} \frac{dy}{dx} = log(a) [2( {sin}^{ - 1} (x) \times \dfrac{1}{ \sqrt{1 - {x}^{2} } } ] }}$

$\\ \implies { \bold{ \dfrac{dy}{dx} = y [ log(a) \{2( {sin}^{ - 1} (x) \times \dfrac{1}{ \sqrt{1 - {x}^{2} } } \} ] }}$

$\\ \implies { \bold{ \dfrac{dy}{dx} = [{a}^{ (\sin {}^{ - 1} x) {}^{2} } ] [ log(a) \{2( {sin}^{ - 1} (x) \times \dfrac{1}{ \sqrt{1 - {x}^{2} } } \} ] }}$

$\\ \implies \large { \red{ \boxed{ \bold{ \dfrac{dy}{dx} = [{a}^{ (\sin {}^{ - 1} x) {}^{2} } ] [ \dfrac{2 log(a). {sin}^{ - 1} (x) }{ \sqrt{1 - {x}^{2} } } ] }}}}$

$\rule{210}{2}$

USED FORMULA :–

$\\ { \green{ \bold{(1) \: \: \frac{d[kf(x)] }{dx} = k \frac{d [f(x)] }{dx} \: \: \: \: [where \: \: k \: \: is \: \: constant] }}}$

$\\ { \green{ \bold{(2) \: \: \frac{d[ {sin}^{ - 1}(x) ] }{dx} = \frac{1}{ \sqrt{1 - {x}^{2} } } }}}$

$\\ { \green{ \bold{(3) \: \: \frac{d ( {x}^{n} ) }{dx} = n {x}^{n - 1} }}}$

$\\ { \green{ \bold{(4) \: \: \frac{d [ log(x) ]}{dx} = \frac{1}{x} }}}$

$\rule{210}{2}$

Answer 2

${\underline{\sf{Given\: Function}}}$

$\sf{y={a}^{ (\sin {}^{-1} x) {}^{2}} }$

${\underline{\sf{To\:Find}}}$

${\sf{\dfrac{dy}{dx}}}$

${\underline{\sf{Solution}}}$

We have , $\bf y={a}^{ ( {sin}^{-1} x){}^{2}} }$

Now take log on both sides

$\sf{\log\:y=\log\:{a}^{ (\sin {}^{-1} x) {}^{2}}}$

$\sf{\log\:y=(\sin {}^{-1} x) {}^{2}\log a }$

Now ,Differentiate with respect to x , by chain rule .

$\implies \bf \frac{1}{y} \times \frac{dy}{dx} = (log\:a \times 2 sin{}^{-1}x) \times \frac{1}{\sqrt{1-x{}^{2}}}$

$\implies \bf \frac{dy}{dx}= 2y \times log\:a \times \frac{sin{}^{-1}x }{ \sqrt{1-x{}^{2} }}$

$\implies \bf \frac{dy}{dx} =2 {a}^{ (\sin {}^{-1} x) {}^{2}} \times log\:a \times \frac{sin{}^{-1}x }{ \sqrt{1-x{}^{2} }}$

$\rule{200}2$

Formula's used :

$\sf{ 1)\dfrac{\log\:x}{dx}=\dfrac{1}{x}}$

$\bf 2)\: \dfrac{sin^{-1}x}{dx} = \dfrac{1}{(\sqrt{1- {x}^{2}})}$

Chain rule :

Let y = f(t) ,t= g(u) and u = m(x) ,then

$\sf \dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$