How would you find the derivative using chain rule when you have ln (x+5)^3?

Question

I understand that $u=(x+5)^3$ and $u'=3(x+5)^2$ but then $v=ln(u)$ and $v'=\frac{u'}{u}$.

Because $v'=\frac{u'}{u}$, is it necessary to plug undifferentiated $u$ into the chain rule formula twice?

Answer 1

No, you’ve got it. The derivative of $\ln(u)$ is just $\frac{u’}{u}$, with no doubling up.

Answer 2

You are on the right track. There's no need to add another $u$.

Chain rule: $$\frac{dv}{dx}=\frac{dv}{du}\cdot\frac{du}{dx}$$

Setting $$u=(x+5)^3$$ and $$v=\ln(x+5)^3=\ln u$$

We have $$\frac{dv}{dx}=\frac{d}{du}(\ln u)\cdot\frac{d}{dx}(x+5)^3$$

So $$\frac{dv}{dx}=\frac{1}{u}\cdot 3(x+5)^2=\frac{3(x+5)^2}{(x+5)^3}=3(x+5)^{-1}$$

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Actually an easier way is to write $$v=\ln(x+5)^3=3\ln(x+5)$$

Immediately, you can see that $$v'(x)=\frac{3}{x+5}$$

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Note:

In general, the relation $$\frac{d}{dx}\ln u(x)=\frac{u'(x)}{u(x)}$$ is already derived via chain rule.