Question

Implicit differentiation (steps needed)

${2}^{x} + {x}^{y} = 1$
Find dy/dx​

Answer 1

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

The given function is

$\rm :\longmapsto\: {2}^{x} + {x}^{y} = 1$

Let we assume that,

$\rm :\longmapsto\:u = {2}^{x}$

and

$\rm :\longmapsto\:v= {x}^{y}$

So, given function can be rewritten as

$\purple{\rm :\longmapsto\: \bf{ \: u + v = 1}}$

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

$\purple{\rm :\longmapsto\: \bf{ \: \dfrac{d}{dx}u + \dfrac{d}{dx}v = \dfrac{d}{dx}1}}$

$\purple{\rm :\longmapsto\: \bf{ \: \dfrac{du}{dx} + \dfrac{dv}{dx} = 0 - - - - (1)}}$

Now, Consider

$\rm :\longmapsto\:u = {2}^{x}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}u = \dfrac{d}{dx} {2}^{x}$

$\red{\bf :\longmapsto\:\dfrac{du}{dx} = {2}^{x} \: log2 - - - - (2)}$

Now, Consider

$\rm :\longmapsto\:v= {x}^{y}$

Taking log on both sides, we get

$\rm :\longmapsto\:logv= log{x}^{y}$

$\rm :\longmapsto\:logv= y \: logx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logv=\dfrac{d}{dx} y \: logx$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = y\dfrac{d}{dx}logx \: + \: logx\dfrac{d}{dx}y$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = y \times \dfrac{1}{x} \: + \: logx\dfrac{dy}{dx}$

$\rm :\longmapsto\: \dfrac{dv}{dx} =v\bigg[\dfrac{y}{x} + logx\dfrac{dy}{dx} \bigg]$

$\rm :\longmapsto\: \dfrac{dv}{dx} = {x}^{y} \bigg[\dfrac{y}{x} + logx\dfrac{dy}{dx} \bigg]$

$\red{\rm :\longmapsto\:\dfrac{dv}{dx} = {x}^{y - 1}y + {x}^{y}logx\dfrac{dy}{dx} - - - - (3)}$

On substituting equation (2) and (3), in equation (1), we get

$\rm :\longmapsto\: {2}^{x} \: log2 + {x}^{y - 1}y + {x}^{y}logx\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\: {2}^{x} \: log2 + {x}^{y - 1}y = - {x}^{y}logx\dfrac{dy}{dx}$

$\bf\implies \:\dfrac{dy}{dx} = - \: \dfrac{ {2}^{x}log2 + {x}^{y - 1}y}{ {x}^{y}logx}$

Additional Information :-

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