Question

2. Find dy/dx when (cos x) ^y = (sin y) ^x​

Answer 1

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

Given function is

$\rm \: {(cosx)}^{y} = {(siny)}^{x}$

Taking log on both sides, we get

$\rm \: log {(cosx)}^{y} = log{(siny)}^{x}$

$\rm \: y \: log(cosx) = x \: log(siny)$

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

$\rm \:\dfrac{d}{dx}[ y \: log(cosx)] =\dfrac{d}{dx}[ x \: log(siny)]$

We know,

$\boxed{ \rm{ \:\dfrac{d}{dx}(u.v) \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$

So, using this, we get

$\rm \: y\dfrac{d}{dx}log(cosx) + log(cosx)\dfrac{d}{dx}y = x\dfrac{d}{dx}log(siny) + log(siny)\dfrac{d}{dx}x$

We know,

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

So, using this result, we get

$\rm \: y\dfrac{1}{(cosx)}\dfrac{d}{dx}(cosx) + log(cosx)\dfrac{dy}{dx} = x\dfrac{1}{(siny)}\dfrac{d}{dx}(siny) + log(siny)$

$\rm \: - y\dfrac{1}{(cosx)}(sinx) + log(cosx)\dfrac{dy}{dx} = \dfrac{x}{(siny)}(cosy)\dfrac{dy}{dx} + log(siny)$

$\rm \: - y(tanx) + log(cosx)\dfrac{dy}{dx} = x(coty)\dfrac{dy}{dx} + log(siny)$

$\rm \:log(cosx)\dfrac{dy}{dx} - x(coty)\dfrac{dy}{dx} = log(siny) + y(tanx)$

$\rm \:\bigg(log(cosx) - x(coty)\bigg)\dfrac{dy}{dx} = log(siny) + y(tanx)$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{log(siny) + y(tanx)}{log(cosx) - x(coty)}$

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

Additional Information :-

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

Answer 2

$\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} =$

Given function is

$\begin{gathered}\rm \: {(cosx)}^{y} = {(siny)}^{x} \\ \end{gathered} </p><p>(cosx) </p><p>y</p><p> =(siny) </p><p>x</p><p>$

Taking log on both sides, we get

$\begin{gathered}\rm \: log {(cosx)}^{y} = log{(siny)}^{x} \\ \end{gathered} </p><p>log(cosx) </p><p>y</p><p> =log(siny) </p><p>x$

$\begin{gathered}\rm \: y \: log(cosx) = x \: log(siny) \\ \end{gathered} </p><p>ylog(cosx)=xlog(siny)$

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

$\begin{gathered}\rm \:\dfrac{d}{dx}[ y \: log(cosx)] =\dfrac{d}{dx}[ x \: log(siny)] \\ \end{gathered} </p><p>dx</p><p>d</p><p></p><p> [ylog(cosx)]= </p><p>dx</p><p>d</p><p></p><p> [xlog(siny)]$

We know,

$\begin{gathered}\boxed{ \rm{ \:\dfrac{d}{dx}(u.v) \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }} \\ \end{gathered} </p><p>dx</p><p>d</p><p></p><p> (u.v)=u </p><p>dx</p><p>d</p><p></p><p> v+v </p><p>dx</p><p>d</p><p></p><p> u$

So, using this, we get

$\begin{gathered}\rm \: y\dfrac{d}{dx}log(cosx) + log(cosx)\dfrac{d}{dx}y = x\dfrac{d}{dx}log(siny) + log(siny)\dfrac{d}{dx}x \\ \end{gathered} </p><p>y </p><p>dx</p><p>d</p><p></p><p> log(cosx)+log(cosx) </p><p>dx</p><p>d</p><p></p><p> y=x </p><p>dx</p><p>d</p><p></p><p> log(siny)+log(siny) </p><p>dx</p><p>d</p><p></p><p> x$

We know,

$\begin{gathered}\boxed{ \rm{ \:\dfrac{d}{dx}logx = \frac{1}{x} \: }} \\ \end{gathered} </p><p>dx</p><p>d</p><p></p><p> logx= </p><p>x</p><p>1$

So, using this result, we get

$\begin{gathered}\rm \: y\dfrac{1}{(cosx)}\dfrac{d}{dx}(cosx) + log(cosx)\dfrac{dy}{dx} = x\dfrac{1}{(siny)}\dfrac{d}{dx}(siny) + log(siny) \\ \end{gathered} </p><p>y </p><p>(cosx)</p><p>1</p><p></p><p> </p><p>dx</p><p>d</p><p></p><p> (cosx)+log(cosx) </p><p>dx</p><p>dy</p><p></p><p> =x </p><p>(siny)</p><p>1</p><p></p><p> </p><p>dx</p><p>d</p><p></p><p> (siny)+log(siny)</p><p>$

$\begin{gathered}\rm \: - y\dfrac{1}{(cosx)}(sinx) + log(cosx)\dfrac{dy}{dx} = \dfrac{x}{(siny)}(cosy)\dfrac{dy}{dx} + log(siny) \\ \end{gathered} </p><p>−y </p><p>(cosx)</p><p>1</p><p></p><p> (sinx)+log(cosx) </p><p>dx</p><p>dy</p><p></p><p> = </p><p>(siny)</p><p>x</p><p></p><p> (cosy) </p><p>dx</p><p>dy</p><p></p><p> +log(siny)$

$\begin{gathered}\rm \: - y(tanx) + log(cosx)\dfrac{dy}{dx} = x(coty)\dfrac{dy}{dx} + log(siny) \\ \end{gathered} </p><p>−y(tanx)+log(cosx) </p><p>dx</p><p>dy</p><p></p><p> =x(coty) </p><p>dx</p><p>dy</p><p></p><p> +log(siny)</p><p>$

$\begin{gathered}\rm \:log(cosx)\dfrac{dy}{dx} - x(coty)\dfrac{dy}{dx} = log(siny) + y(tanx) \\ \end{gathered} </p><p>log(cosx) </p><p>dx</p><p>dy</p><p></p><p> −x(coty) </p><p>dx</p><p>dy</p><p></p><p> =log(siny)+y(tanx)</p><p>$

$\begin{gathered}\rm \:\bigg(log(cosx) - x(coty)\bigg)\dfrac{dy}{dx} = log(siny) + y(tanx) \\ \end{gathered} </p><p>(log(cosx)−x(coty)) </p><p>dx</p><p>dy</p><p></p><p> =log(siny)+y(tanx)</p><p></p><p>$

$\begin{gathered}\bf\implies \:\dfrac{dy}{dx} = \dfrac{log(siny) + y(tanx)}{log(cosx) - x(coty)} \\ \end{gathered} </p><p>⟹ </p><p>dx</p><p>dy</p><p></p><p> = </p><p>log(cosx)−x(coty)</p><p>log(siny)+y(tanx)</p><p></p><p> </p><p></p><p> </p><p></p><p>\rule{190pt}{2pt}$

Additional Information :-

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