Question

Find $\frac{dy}{dx}$ of the given function : $(cosx{)}^y=(cosy{)}^x$.

• A. $-yco{s}^{y}xsinx$ = $cos{x}^{y}logy$
• B. $-yco{s}^{y-1}xsinx$ = $cos{y}^{x}logcosy$
  
• C. $-yco{s}^{x-1}ysinx$ = $cos{y}^{x}logcosy$
• D. $-xco{s}^{x-1}xsinx$ = $cos{y}^{x}logcosy$

Answer 1

The correct option is B

$-yco{s}^{y-1}xsinx$ = $cos{y}^{x}logcosy$

$(cosx{)}^y=(cosy{)}^x$

Now, differentiating with respect to x

$y{\cos }^{y-1}x.(-\sin x)=\left(\cos y{)}^x\log \right(\cos y)$

$-yco{s}^{y-1}x.sinx$ = $cos{y}^{x}log\left(cosy\right)$