differentiate the following function with respect to x

(log x)^ cos x

y=logx^cosx

^{taking log on both sids}

logy=cosx logx

diff. w. r. to x , we get

(1/y)dy/dx=cosx.(1/x)+logx(-sinx)

dy/dx=logx^cosx[(cosx)/x –sinx .logx]

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty