How do you find the derivative of #y=x^cosx#?

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Answer 1 · 2016-08-20T16:08:55

#dy/dx = x^(cosx)(-sinxlnx + cosx/x)#

Take the natural logarithm of both sides:

#lny = ln(x^(cosx))#

#lny = cosx(ln(x))#

Now, take the derivative of both sides.

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

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

#dy/dx = x^(cosx)(-sinxlnx + cosx/x)#

Hopefully this helps!