How do you differentiate #f(x)=lnx^2*e^(x^3-1)# using the product rule?

1 Answer
Nov 5, 2016

Answer:

Use the chain rule in conjunction with the product rule.

Explanation:

We can start by using the chain rule to take the derivative of our first term, #lnx^2#. We take the derivative of the "outside," then multiply it by the derivative of the "inside." Remember that the derivative of the natural log is #1/x#.

Thus, we get #(2ln(x))/x#.

By the product rule, since we've taken the derivative of the first term, we multiply by the second term, which we leave alone.

This gives us, so far: #(2ln(x))/x*e^(x^3-1)#.

Now we do the same thing to the other term. We take the derivative of #e^(x^3-1)# using the chain rule, and multiply by the original of the first term, giving us #3lnx^2e^(x^3-1)x^2#.

By the product rule, we then add these two resulting terms together.

#(2ln(x)e^(x^3-1))/x + 3lnx^2e^(x^3-1)x^2#.

This can be further simplified as:

#(e^(x^3-1)ln(x)(3x^3ln(x)+2))/x#