How do you find the derivative of #y = x^(cos x)#?

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Answer 1 · 2017-02-18T18:47:54

#dy/dx=x^cosx((cosx)/x-lnxsinx)#

Take the natural log of both sides and move the cos in front of the natural log.

#lny=(cosx)lnx#

Use implicit differentiation. You have to take the product rule of the right side.

#(1/y)(dy/dx)=(cosx)(1/x)+(lnx)(-sinx)#

#(1/y)(dy/dx)=(cosx)/x-lnxsinx#

Solve for #dy/dx#

#dy/dx=y((cosx)/x-lnxsinx)#

Plug in #x^cosx# for y

#dy/dx=x^cosx((cosx)/x-lnxsinx)#