How do you find the derivative of y=cosln(4x^3)?

1 Answer
Feb 5, 2017

(dy)/(dx)=-(3sin(ln(4x^3)))/x

Explanation:

Using chain rule, let u=ln(4x^3)

So y=cosu

(dy)/(dx)=(dy)/(du)*(du)/(dx)

=-sinu*d/(dx)ln(4x^3)

Now we have to derive ln(4x^3) using chain rule as well.

Let y=lnu, u=4x^3

(dy)/(dx)=d/(du)lnu*d/(dx)4x^3

=1/u*12x^2

=(12x^2)/(4x^3)

=3/x

:. d/(dx)ln(4x^3)=-sin(ln(4x^3))*3/x

=-(3sin(ln(4x^3)))/x