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.
Finding the derivatives of logarithms
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calculus
2 Answers
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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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0@Jordan: Since $y$ changes as $x$ changes, we treat $y$ as a function of $x$. Since we don't know what $y$ is, we have to "leave it indicated" as $y'$; otherwise, I think you are on the right track. – 2011-10-01
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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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1Hate 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. – 2011-10-02