How do you differentiate #f(x)=ln2x * sin4x# using the product rule?

1 Answer
Jun 1, 2016

Answer:

f'(x)=(2sin(4x))/2x + 4cos(4x)ln2x

Explanation:

The product rule says you take the derivative of your first function and multiply it by the second and add it to the product of the derivative of your second function and your original first function.

We start with taking the first derivative of function #ln2x#, which is #2/2x#(don't forget to chain rule). we then multiply this by our second function, #sin(4x)#, which results in the first piece #2sin(2x)/2x#.

We then take the derivative of our second function, #sin(4x)#, which is #4cos(4x)#(again don't forget to chain rule). Then multiply this by the first function to get #4cos(4x)ln2x#.

Lastly add them up to get the final answer #f'(x)= (2sin(4x))/2x + 4cos(4x)ln2x#. Hope this helped.