How do you differentiate #y=x^cosx#?

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Answer 1 · 2015-06-06T23:26:07

#y=x^cos(x)#

Take logarithm of each side.
#ln(y)=ln(x^cos(x))#

#ln(y)=cos(x)ln(x)#

Take derivative of each side

#d/dx[ln(y)]=d/dx[cos(x)ln(x)]#

#1/ydy/dx=-sin(x)ln(x)+cos(x)/x#

#dy/dx=y(-sin(x)ln(x)+cos(x)/x )#

Substitute #y#

#dy/dx=x^cos(x)(-sin(x)ln(x)+cos(x)/x )#