Question

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

Answer 1

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

Explanation:

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`