Question

y^x+x^y+x^x=a^n.finddy/dx​

Answer 1

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

Given expression is

$\rm \: {y}^{x} + {x}^{y} + {x}^{x} = {a}^{n}$

Let assume that,

$\rm \: u + v + w = {a}^{n}$

where,

$\rm \: u = {y}^{x}$

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

$\rm \: w = {x}^{x}$

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

$\rm \: \dfrac{d}{dx} (u + v + w) = \dfrac{d}{dx} {a}^{n}$

$\rm \: \dfrac{d}{dx}u + \dfrac{d}{dx}v + \dfrac{dy}{dx}w = 0$

$\rm\implies \:\dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx} = 0 - - - - (1)$

Now, Consider

$\rm \: u = {y}^{x}$

On taking log on both sides, we get

$\rm \: logu = log{y}^{x}$

$\rm \: logu = xlogy$

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

$\rm \: \dfrac{d}{dx}logu = \dfrac{d}{dx}xlogy$

$\rm \: \dfrac{1}{u}\dfrac{du}{dx} = x\dfrac{d}{dx}logy + logy\dfrac{d}{dx}x$

$\rm \: \dfrac{1}{u}\dfrac{du}{dx} = x\dfrac{1}{y}\dfrac{dy}{dx}+ logy \times 1$

$\rm \: \dfrac{du}{dx} = u\bigg(\dfrac{x}{y}\dfrac{dy}{dx}+ logy\bigg)$

$\rm\implies \:\rm \: \dfrac{du}{dx} = {y}^{x} \bigg(\dfrac{x}{y}\dfrac{dy}{dx}+ logy\bigg) - - - - (2)$

Now, Consider

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

On taking log on both sides, we get

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

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

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

$\rm \: \dfrac{d}{dx}logv = \dfrac{d}{dx} \: y \: log{x}$

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

$\rm \: \dfrac{dv}{dx} =v\bigg( y\dfrac{1}{x} + logx\dfrac{dy}{dx}\bigg)$

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

Now, Consider

$\rm \: w = {x}^{x}$

On taking log on both sides, we get

$\rm \: logw = log{x}^{x}$

$\rm \: logw = xlog{x}$

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

$\rm \: \dfrac{d}{dx}logw = \dfrac{d}{dx} xlog{x}$

$\rm \: \dfrac{1}{w}\dfrac{dw}{dx} = x\dfrac{d}{dx}logx + logx\dfrac{d}{dx}x$

$\rm \: \dfrac{dw}{dx} = w\bigg(x \times \dfrac{1}{x} + logx \times 1 \bigg)$

$\rm\implies \:\rm \: \dfrac{dw}{dx} = {x}^{x} \bigg(1 + logx \bigg) - - - - (4)$

So, on substituting the values from equation (2), (3) and (4) in equation (1), we get

$\rm \: {y}^{x} \bigg(\dfrac{x}{y}\dfrac{dy}{dx}+ logy\bigg) + {x}^{y} \bigg( y\dfrac{1}{x} + logx\dfrac{dy}{dx}\bigg) + {x}^{x}(1 + logx) = 0$

$\rm \: x{y}^{x - 1}\dfrac{dy}{dx}+ {y}^{x} logy +y {x}^{y - 1} + {x}^{y} logx\dfrac{dy}{dx} + {x}^{x}(1 + logx) = 0$

$\rm \: x{y}^{x - 1}\dfrac{dy}{dx} + {x}^{y} logx\dfrac{dy}{dx} = - \bigg({y}^{x} logy +y {x}^{y - 1} + {x}^{x}(1 + logx)\bigg)$

$\rm \: \bigg(x{y}^{x - 1} + {x}^{y} logx\bigg)\dfrac{dy}{dx} = - \bigg({y}^{x} logy +y {x}^{y - 1} + {x}^{x}(1 + logx)\bigg)$

$\rm \: \dfrac{dy}{dx} = \: - \: \dfrac{\bigg({y}^{x} logy +y {x}^{y - 1} + {x}^{x}(1 + logx)\bigg)}{\bigg(x{y}^{x - 1} + {x}^{y} logx\bigg)}$

Hence,

$\\ \boxed{\sf{  \:\rm \: \dfrac{dy}{dx} = \: - \: \dfrac{\bigg({y}^{x} logy +y {x}^{y - 1} + {x}^{x}(1 + logx)\bigg)}{\bigg(x{y}^{x - 1} + {x}^{y} logx\bigg)}}}$

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Formulae Used :-

$\boxed{\sf{  \: \: \dfrac{d}{dx}logx \: = \: \frac{1}{x} \: }}$

$\boxed{\sf{  \:\dfrac{d}{dx}x \: = \: 1 \: }}$

$\boxed{\sf{  \:\dfrac{d}{dx}k \: = \: 0 \: }}$

$\boxed{\sf{  \:\dfrac{d}{dx}uv \: = \: u \: \dfrac{d}{dx}v \: + \: v \: \dfrac{d}{dx}u \: \: }}$

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Additional Information :-

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