Question

If $(\cos x{)}^y=(\sin y{)}^x$, then $\frac{dy}{dx}=$

• A. $\frac{\log \left(\sin y\right)+y\tan x}{\log \left(\cos x\right)-x\cot y}$
• B. $\frac{\log \left(\sin y\right)-y\tan x}{\log \left(\cos x\right)+x\cot y}$
• C. $\frac{\log \left(\sin y\right)}{\log \left(\cos x\right)}$
• D. $\frac{\log \left(\cos x\right)}{\log \left(\sin y\right)}$

Answer 1

The correct option is B $\frac{\log \left(\sin y\right)-y\tan x}{\log \left(\cos x\right)+x\cot y}$

Given,

$(\cos x{)}^y=(\sin y{)}^x$

Now taking logarithm with respect to the base $e$ we get,

$y\log \left(\cos x\right)=x\log \left(\sin y\right)$

Now differentiating both sides with respect to $x$ we get,

$\log \left(\cos x\right)\frac{dy}{dx}+y.\frac{(-\sin x)}{\cos x}$$=\log \left(\sin y\right)+x.\frac{\left(\cos y\right)}{\sin y}\frac{dy}{dx}$

or, $\left(\log \right(\cos x)-x\cot y)\frac{dy}{dx}=\log \left(\sin y\right)+y\tan x$

or, $\frac{dy}{dx}=\frac{\log \left(\sin y\right)+y\tan x}{\log \left(\cos x\right)-y\cot x}$