How do you differentiate # g(x) = x^cosx #?

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Answer 1 · 2016-06-03T01:13:26

#g'(x)=(x^cos(x)(cos(x)-xsin(x)ln(x)))/x#

Let

#y=x^cos(x)#

take the natural logarithm of both sides.

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

Simplify the right-hand side using the rule: #ln(a^b)=b*ln(a)#

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

Differentiate both sides. The left-hand side will use the chain rule, and the right hand side the product rule.

#1/y(dy/dx)=-sin(x)*ln(x)+cos(x)*1/x#

Simplifying the right-hand side:

#1/y(dy/dx)=(cos(x)-xsin(x)ln(x))/x#

Solve for #dy/dx# by multiplying both sides by #y#, or #x^cos(x)#.

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