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I'm asked to prove that the derivative of $$\frac{\ln{x}}{\ln{a}}$$ is $$\frac{1}{x\cdot\ln{a}}$$

My attempt:

$$\frac{d}{dx}(\frac{\ln{x}}{\ln{a}}) = \frac{\frac{1}{x}\cdot \ln{a} - \frac{1}{a}\cdot \ln{x}}{(\ln{a})^2}$$ which is $$\frac{1}{x\cdot\ln{a}} - \frac{\ln{x}}{a\cdot (\ln{a})^2}$$

which is not equal to $$\frac{1}{x\cdot\ln{a}}$$

What am I doing wrong?

user359689
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    Here $a$ is a constant, so the chain rule need not be used. You can just pull out the term $1/\ln a$ from $d/dx(\ln x / \ln a)$. –  Aug 06 '16 at 11:34
  • If you're using the quotient rule, the derivative of $\ln a$ is $0$, not $\frac{1}{a}$. – D_S Aug 06 '16 at 12:29

1 Answers1

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In your usage of the fraction derivative rule $$\left(\frac uv\right)'=\frac{u'v-v'u}{v^2}$$ please have a problem addressed by @Brahadeesh in the comments:

Here $a$ is a constant.

That is, please notice, however your denominator and the $u'v$ part of your numerator are perfectly fine, the $v'u$ part had been evaluated incorrectly.

Having the quoted part of the comment in mind, can you please now try evaluating $$\left(\ln a\right)'\ln x$$ again, where the differentiation is meant to be with respect to $x$, as per the original problem?

Whenever you find and correct your problem, you can also try a different approach (also suggested by @Brahadeesh), which would give the required answer with less evaluation needed (not implying your original approach is incorrect). That goes as follows.

Please notice, $$\frac{\ln x}{\ln a}=\frac1{\ln a}\ln x;$$ now apply the identity $$\left(kf(x)\right)'=kf'(x)$$ with $k=\dfrac1{\ln a}$: $$\left(\frac{\ln x}{\ln a}\right)'=\frac1{\ln a}\left(\ln x\right)';$$ evaluate further to arrive at the required answer (the prime, again, denotes differentiation with respect to $x$, as per the original problem).

dbanet
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