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

Nov 5, 2016

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, $\ln {x}^{2}$. 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 $\frac{1}{x}$.

Thus, we get $\frac{2 \ln \left(x\right)}{x}$.

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: $\frac{2 \ln \left(x\right)}{x} \cdot {e}^{{x}^{3} - 1}$.

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 $3 \ln {x}^{2} {e}^{{x}^{3} - 1} {x}^{2}$.

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

$\frac{2 \ln \left(x\right) {e}^{{x}^{3} - 1}}{x} + 3 \ln {x}^{2} {e}^{{x}^{3} - 1} {x}^{2}$.

This can be further simplified as:

$\frac{{e}^{{x}^{3} - 1} \ln \left(x\right) \left(3 {x}^{3} \ln \left(x\right) + 2\right)}{x}$