Question

differentiation ofcos(x^x)
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Answer 1

Answer:

dy/dx=y.log(cosx)-tanx

Step-by-step explanation:

y=cosx^x  

consider log

logy=log(cosx^x)

differntiation of log is 1/x

1/y.dy/dx=x.log(cosx)                             {as logm^n=nlogm}

1/y.dy/dx=1.log(cosx)+x.1/cosx.d/dx(cosx)         {product rule in rhs }

1/y.dy/dx=log(cosx)+1/cosx.(-sinx)                    derivative of cosx=-sinx

1/y.dy/dx=log(cosx)-sinx/cosx

dy/dx=y.log(cosx)-tanx

hope it helps...if dont understood...tell me