Question

differentiate w.r.t 'x',
${(sinx)}^{x}$
+
${sin}^{? - 1} \sqrt{x}$
​

Answer 1

ANSWER:

$y' = {(sinx)}^{x-1} \left(x \: cos \ x+sin \ x \: log(sin \ x)\right)+ \dfrac{1}{ \sqrt{1 - x}}$

GIVEN:

${(sinx)}^{x} + {sin}^{ - 1} \sqrt{x}$

TO DIFFERENTIATE:

${(sinx)}^{x} + {sin}^{ - 1} \sqrt{x}$

FORMULAE USED:

$\dfrac{d}{dx} y = y'$

$\log \ x^y = y \log \ x$

$\dfrac{d}{dx} log \: x = \dfrac{1}{x}$

$\dfrac{d}{dx}uv = uv'+ vu'$

$\dfrac{ {x}^{y} }{{x}^{z} } = {x}^{y - z}$

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

EXPLANATION:

$Let \ y = {(sinx)}^{x} + {sin}^{ - 1} \sqrt{x}$

$Let \ A = {(sinx)}^{x}$

$Let \ B ={sin}^{ - 1} \sqrt{x}$

y = A + B

y' = A' + B'

$Take \ A = {(sinx)}^{x}$

Multiply by log on both sides

$log\ A = log{(sinx)}^{x}$

$log\ A = x \ log(sinx)$

Differentiate w.r.t x

$\dfrac{d}{dx}log \ A = x\dfrac{d}{dx}log(sin \ x)+log(sin\ x)\dfrac{d}{dx}( x)$

$\dfrac{1}{A}(A' )= x \dfrac{1}{sin \ x}(cos \ x)+log(sin \ x)$

$A'= A \left(\dfrac{x \: cos \ x+sin \ x \: log(sin \ x)}{sin \ x} \right)$

$A'= {(sinx)}^{x} \left(\dfrac{x \: cos \ x+sin \ x \: log(sin \ x)}{sin \ x} \right)$

$A'= {(sinx)}^{x - 1} \left(x \: cos \ x+sin \ x \: log(sin \ x) \right)$

$Take \ B ={sin}^{ - 1} \sqrt{x}$

Differentiate w.r.t x

$B' = \dfrac{1}{ \sqrt{1 -{ (\sqrt{x}})^{2} } }$

$B' = \dfrac{1}{ \sqrt{1 - x} }$

y' = A' + B'

$y' = {(sinx)}^{x-1} \left(x \: cos \ x+sin \ x \: log(sin \ x)\right)+ \dfrac{1}{ \sqrt{1 - x}}$]