How do you differentiate g(x) = x^cosx ?
1 Answer
Jun 3, 2016
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(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=(x^cos(x)(cos(x)-xsin(x)ln(x)))/x