# Differentiating $\ln(x^3)$ using the chain rule

Because $\ln a^b=b\ln a$, $\frac{d}{dx} \ln x^3$ should equal $\frac{3}{x}$. However, if you apply the chain rule, I believe you'd get $(3x^2)\frac{x^3}{x}$. Where is my reasoning wrong?

• why do you have a x^3 in the numerator – Saketh Malyala May 7 '17 at 3:50

If you apply the chain rule you should have $\dfrac{1}{x^3} \cdot 3x^2 = \dfrac{3}{x}$.

• That makes sense; thank you! But why is it that the inner function $(x^3)$ is not kept when differentiating $ln(x^3)$. For example $\frac{d}{dx}\cos(x^2) = 2x\sin(x^2)$, right? – hyperdo May 7 '17 at 3:53
• It is kept. remember d/dx ln(u) = 1/u * dx/du. we keep the inner function and move it to the bottom, and multiply by dx/du, which is 3x^2 in this case – Saketh Malyala May 7 '17 at 3:54
• Oh, I see. Thanks so much for your help! I'll make it as accepted as soon as it will allow me. :) – hyperdo May 7 '17 at 3:56

By the chain rule, $\frac{d}{dx} \ln u=\frac1uu'$.

In your case, $u=x^3$, so $u'=3x^2$.

Thus, $\frac{d}{dx} \ln x^3=\frac{1}{x^3}3x^2=\frac3x$.

The error in your reasoning was keeping the $x^3$ around and dividing by $x$ instead of $x^3$.