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I am attempting to find the derivative of $\log_{10} (x^3 + 1)$ I am not too sure to do with this actually, I know the formula states that it should be $1/ (x\ln a)$ but does that mean just plug it in to that or do I need to use the product rule? I am not too sure what to do.

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up vote 2 down vote accepted

The formula for the derivative of a logarithm says $$\frac{d}{du} \log_a(u) = \frac{1}{u\ln(a)}.$$ What you have here is $\log_{10}(x^3+1)$, so you need to use the Chain Rule: $$\frac{d}{dx}\log_{10}(x^3+1) = \frac{1}{(x^3+1)\ln(10)}\left(\frac{d}{dx}(x^3+1)\right).$$

There are no products in your function, so the Product Rule is not "in play".

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...and since it'll come up often, one might want to remember the pattern $\dfrac{f'(x)}{f(x)}$... – J. M. Oct 1 '11 at 23:24
I am a little confused then, why am I using the chain rule if x is just $x^3+1$ why am I using the chain rule also? – user138246 Oct 1 '11 at 23:26
@Jordan: You are using the Chain Rule because this is a composition: it's the composition of the function $f(x)=x^3+1$, and the function $g(u) = \log_{10}(u)$. You always use the Chain Rule when you have to do the derivative of a composition. That's what it's for! Will you be less confused if I change the $x$ to a $u$ in the formula for the derivative of the logarithm? – Arturo Magidin Oct 1 '11 at 23:28
Okay, I was just a little confused by the $1/xlna$ I guess I can't really just use that alone can I? It would only work if it was $log_a(x)$ or something similar, where x is just a single variable or constant? – user138246 Oct 1 '11 at 23:30
@Jordan: So your rate of change relative to my income is 0.1 (10% of whatever my change is). My rate of change relative to my neighbor's income is 0.15 (15% of whatever his change is). But your rate of change relative to my neighbor's income is not 10%, and it's not 15%, it's the product of your change realtive to me, times my change relative to him: $.1\times .15 = .015$, or 1.5%. You don't use the same formula to compute both, because you are trying to figure out the change with respect to different things. – Arturo Magidin Oct 1 '11 at 23:52

Without using chain-rule we can do in this way.

Let $log_{10}(x^3+1) = y \implies x^3+1= 10^y$

Now differentiating on both sides , $3x^2 = (ln10)(10^y)\left(\frac{\displaystyle dy}{\displaystyle dx}\right)$

So $\frac{\displaystyle dy}{\displaystyle dx}= \frac{\displaystyle 3x^2}{\displaystyle (x^3+1)(ln10)}$

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Hate to break it to you, but you are using the Chain Rule: you use it when you claim that the derivative of $10^y$ is $(\ln 10)10^y\frac{dy}{dx}$; that last bit? That's the Chain Rule. – Arturo Magidin Oct 2 '11 at 0:06

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