Question

y=(cosx)^x

Answer 1

do you want to find y'?

Answer 2

yes

Answer 3

have you tried to do this
we have done two

Answer 4

do the same thing again take ln on both sides and differentiate implicitly..

Answer 5

i think this is the right file...

Answer 6

take natural log of both sides
i will walk you through it

Answer 7

\begin{eqnarray*}y' &=& (e^{x \log{(\cos{x})}})' \\
&=& e^{x\log{(\cos{x})}}(x\log{(\cos{x})})' \\
&=& e^{x\log{(\cos{x})}}\left( x\frac{1}{\cos{x}} (-\sin{x}) + \log{(\cos{x})} \right) \\
&=& (\cos{x})^x(\log{(\cos{x})} - x\tan{x}). \end{eqnarray*}

Answer 8

I may have done this wrong but I got y'=(cosx)^x(xln(-sinx)+(cosx/x))