differentiate the following function with respect to x

(sin x)^cos x

let y1=(sinx)^cosx

^{taking log on both sides}

logy1=cosx.logsinx

diff. w. r. to x using product rule,we get

1/y1.dy1/dx=cosx.(1/sinx).cosx+logsinx.(-sinx)

dy1/dx=(sinx)^cosx[cos²x- sin²x.logsinx]/sinx

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty