Question

if y = (sin x)^x + sin^-1 x - 2^sinx, find dy/dx

Answer 1

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

Given that

$\rm :\longmapsto\:y = {(sinx)}^{x} + {sin}^{ - 1}x - {2}^{sinx}$

$\rm :\longmapsto\:Let \: y \: = \: u + v + w - - (1)$

where,

$\red{\rm :\longmapsto\: \:u = {(sinx)}^{x} } \\ \red{\rm :\longmapsto\: \:v = {sin}^{ - 1}x} \\ \red{\rm :\longmapsto\: \: w = {2}^{sinx} } \: \: \:$

On differentiating equation (1), w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}u + \dfrac{d}{dx}v + \dfrac{d}{dx}w$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx} - - - (2)$

Now,.

Consider,

$\rm :\longmapsto\:u = {(sinx)}^{x}$

☆ On taking log both sides, we get

$\rm :\longmapsto\:logu = log {(sinx)}^{x}$

$\rm :\longmapsto\:logu =x \: log {(sinx)}$

$\red{\bigg \{ \because \:log {x}^{y} = y \: logx \bigg \}}$

☆ On differentiate both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}logu =\dfrac{d}{dx}x \: log {(sinx)}$

$\rm :\longmapsto\:\dfrac{1}{u} \dfrac{du}{dx} =x\dfrac{d}{dx}\: log {(sinx)} + log \: sinx \: \dfrac{d}{dx}x$

$\red{\bigg \{ \because \:\dfrac{d}{dx}logx = \dfrac{1}{x} \bigg \}} \\ \red{\bigg \{ \because \: \dfrac{d}{dx}u.v = u\dfrac{d}{dx}v + v\dfrac{d}{dx}u\bigg \}}$

$\rm :\longmapsto\:\dfrac{1}{u} \dfrac{du}{dx} =x \: \dfrac{1}{sinx} \: \dfrac{d}{dx}sinx + log \: sinx \:$

$\red{\bigg \{ \because \:\dfrac{d}{dx}x = 1 \bigg \}}$

$\rm :\longmapsto\:\dfrac{1}{u} \dfrac{du}{dx} =x \: \dfrac{1}{sinx} \: cosx + log \: sinx \:$

$\red{\bigg \{ \because \:\dfrac{d}{dx}sinx = cosx \bigg \}}$

$\rm :\longmapsto\:\dfrac{du}{dx} = u\bigg(x \: cotx \: + \: log \: sinx\bigg)$

$\bf\implies \:\:\dfrac{du}{dx} = {(sinx)}^{x} \bigg(x \: cotx \: + \: log \: sinx\bigg) - - (3)$

Consider,

$\rm :\longmapsto\:v = {sin}^{ - 1} x$

☆ On differentiate both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}v = \dfrac{d}{dx} {sin}^{ - 1} x$

$\bf\implies \:\:\dfrac{dv}{dx}= \dfrac{1}{ \sqrt{1 - {x}^{2} } } - - - (4)$

Consider,

$\rm :\longmapsto\:w = {2}^{sinx}$

☆ On differentiate both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}w = \dfrac{d}{dx} {2}^{sinx}$

$\rm :\longmapsto\:\dfrac{dw}{dx} = {2}^{sinx}log2 \: \dfrac{d}{dx}sinx$

$\red{\bigg \{ \because \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) \bigg \}}$

$\rm :\longmapsto\:\dfrac{dw}{dx} = {2}^{sinx}log2 \: cosx$

$\bf\implies \:\dfrac{dw}{dx} = {2}^{sinx} \: cosx \: log2 - - - (5)$

On substituting all the values from equation (3),(4),(5) in equation (2), we get

${\rm :\mapsto\:\dfrac{dy}{dx} = {(sinx)}^{x} \bigg(x \: cotx \: + \: log \: sinx\bigg) + \dfrac{1}{ \sqrt{1 - {x}^{2}}} + {2}^{sinx} \: cosx \: log2}$

Additional Information :-

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}x = 1}$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}logx = \dfrac{1}{x} }$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}sinx = cosx}$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}cosx = - sinx}$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}tanx = {sec}^{2} x}$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}cotx = { - cosec}^{2} x}$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}secx = secx \: tanx}$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx} {sin}^{ - 1} x = \dfrac{1}{ \sqrt{1 - {x}^{2} } } }$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx} {cos}^{ - 1} x = \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } } }$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx} {tan}^{ - 1}x = \dfrac{1}{1 + {x}^{2} } }$

$\red{\rm :\longmapsto\: \: \dfrac{d}{dx} {cot}^{ - 1}x = \dfrac{ - 1}{1 + {x}^{2} } }$