Question

If

$x = {e}^{cos2t}$

$y = {e}^{sin2t}$

prove that

$\frac{dy}{dx} = - \frac{y logx\: }{x \: logy}$
​

Answer 1

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

Given that,

$\rm \: x = {e}^{cos2t}$

Taking log on both sides, we get

$\rm \: logx = log {e}^{cos2t}$

$\rm \: logx = cos2t \: loge$

$\rm \: logx = cos2t - - - (1)$

Also, given that

$\rm \: y = {e}^{sin2t}$

Taking log on both sides, we get

$\rm \: logy = log{e}^{sin2t}$

$\rm \: logy = sin2t \: loge$

$\rm \: logy = sin2t - - - (2)$

On squaring equation (1) and (2) and add, we get

$\rm \: {(logx)}^{2} + {(logy)}^{2} = {cos}^{2}2t + {sin}^{2} 2t$

$\rm \: {(logx)}^{2} + {(logy)}^{2} = 1$

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

$\rm \: \dfrac{d}{dx}\bigg[{(logx)}^{2} + {(logy)}^{2} \bigg]= \dfrac{d}{dx}1$

$\rm \: \dfrac{d}{dx} {(logx)}^{2} + \dfrac{d}{dx} {(logy)}^{2} = 0$

$\rm \: 2logx\dfrac{d}{dx}logx + 2logy\dfrac{d}{dx}logy = 0$

$\rm \: 2logx \times \frac{1}{x} + 2logy \times \frac{1}{y}\dfrac{dy}{dx} = 0$

$\rm \: 2\bigg(logx \times \frac{1}{x} + logy \times \frac{1}{y}\dfrac{dy}{dx}\bigg)= 0$

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

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

$\rm\implies \:\boxed{ \rm{ \:\dfrac{dy}{dx} \: = \: - \: \frac{y \: logx}{x \: logy} \: \: }}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: \: }}$

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

$\rule{190pt}{2pt}$

Additional Information :-

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