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

Question

Ezra Adams · Accepted Answer

Use the chain rule in conjunction with the product rule.

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#. We multiply by the second term, which we leave alone, according to the product rule since we have already taken the derivative of the first term. 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#. We then sum these two resultant terms together using the product rule. #(2ln(x)e^(x^3-1))/x + 3lnx^2e^(x^3-1)x^2#. This can be made even simpler as: #(e^(x^3-1)ln(x)(3x^3ln(x)+2))/x#

Adam Adkins · Answer

undefined To differentiate $ f(x) = \ln(x^2) \cdot e^{x^3 - 1} $ using the product rule:

$ f'(x) = (ln(x^2))' \cdot e^{x^3 - 1} + ln(x^2) \cdot (e^{x^3 - 1})' $

Using the chain rule and the derivative of the natural logarithm function, and the derivative of the exponential function:

$ f'(x) = (\frac{d}{dx}(ln(x^2))) \cdot e^{x^3 - 1} + ln(x^2) \cdot \frac{d}{dx}(e^{x^3 - 1}) $

$ f'(x) = (\frac{1}{x^2} \cdot 2x) \cdot e^{x^3 - 1} + ln(x^2) \cdot e^{x^3 - 1} \cdot (3x^2) $

$ f'(x) = \frac{2}{x} \cdot e^{x^3 - 1} + 3x^2 \cdot ln(x^2) \cdot e^{x^3 - 1} $