Question

Differentiate the following function with respect to x

${x}^{y}. {y}^{x} = {a}^{b}$
​

Answer 1

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

Given function is

$\rm \: {x}^{y} . {y}^{x} = {a}^{b}$

Taking log on both sides, we get

$\rm \: log[{x}^{y} . {y}^{x}] = log{a}^{b}$

can be rewritten as

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

$\rm \: y \: logx \: + \: x \: logy \: = \: log {a}^{b}$

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

$\rm \: \dfrac{d}{dx}(y \: logx )\: + \: \dfrac{d}{dx}(x \: logy) \: = \: \dfrac{d}{dx}log {a}^{b}$

Using product rule of differentiation,

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

So, using this, we get

$\rm \: y\dfrac{d}{dx}logx + logx\dfrac{d}{dx}y + x\dfrac{d}{dx}logy + logy\dfrac{d}{dx}x = 0$

We know,

$\boxed{\tt{ \: \: \dfrac{d}{dx}logx = \frac{1}{x}}} \: \: \\ \\ \boxed{\tt{ \: \: \: \dfrac{d}{dx}k \: = \: 0 \ \: \: }}$

So, using this, we get

$\rm \: y \times \dfrac{1}{x} + logx\dfrac{dy}{dx} + x \times \dfrac{1}{y} \dfrac{dy}{dx} + logy \times 1= 0$

$\rm \: \dfrac{y}{x} + logx\dfrac{dy}{dx} + \dfrac{x}{y} \dfrac{dy}{dx} + logy= 0$

$\rm \: logx\dfrac{dy}{dx} + \dfrac{x}{y} \dfrac{dy}{dx}= - logy - \dfrac{y}{x}$

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

$\rm \: \bigg(\dfrac{ylogx + x}{y}\bigg) \dfrac{dy}{dx}= -\bigg(\dfrac{xlogy + y}{x}\bigg)$

$\rm \: \dfrac{dy}{dx}= -\dfrac{y}{x} \bigg(\dfrac{xlogy + y}{ylogx + x}\bigg)$

Hence,

$\rm\implies \:\rm \: \dfrac{dy}{dx}= -\dfrac{y}{x} \bigg(\dfrac{xlogy + y}{ylogx + x}\bigg)$

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ADDITIONAL INFORMATION

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