Multiplication to addition

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Two young mathematicians think about derivatives and logarithms.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, why is the product rule so much harder than the sum rule?
Riley: Ever since 2nd grade, I’ve known that multiplication is harder than addition.
Devyn: I know! I was reading somewhere that a slide-rule somehow turns ‘‘multiplication into addition.’’
Riley: Wow! I wonder how that works?
Devyn: I think it has something to do with logs?
Riley: What? How does this work?

Devyn is right, logarithms are used (and were invented) to convert difficult multiplication problems into simpler addition problems.

Let $f(x) = (x+3) \cdot (x-5) \cdot (x+8)$ . Compute $\ddx f(x)\begin {prompt} = \answer {(x-5)(x+8)+(x+3)(x-5)+(x+3)(x-5)}\end {prompt}$

Now, let’s see what happens if we do the same problem but we take the natural log of both sides first:

$\begin{align*} f(x) &=(x+3) \cdot (x-5) \cdot (x+8)\\ \ln(f(x)) &= \ln((x+3) \cdot (x-5) \cdot (x+8))\\ \ln(f(x)) &=\ln(x+3) + \ln(x-4) + \ln(x+8) \end{align*}$

Now we’ll take the derivative of both sides of the equation. By the chain rule $\ddx \ln (f(x))= \frac {f'(x)}{f(x)}$

Compute $\ddx \ln (x+3) \begin {prompt}=\answer {\frac {1)}{x+3}}\end {prompt}$

Compute $\ddx \ln (x-5) \begin {prompt}=\answer {\frac {1}{x-5}}\end {prompt}$

Compute $\ddx \ln (x+8) \begin {prompt}=\answer {\frac {1}{x+8}}\end {prompt}$

So we have

$\begin{align*} \frac{f'(x)}{f(x)} &= \frac{1}{x+3} + \frac{1}{x-5} + \frac{1}{x+8} \\ f'(x) &= f(x) \left(\frac{1}{x+3} + \frac{1}{x-5} + \frac{1}{x+8}\right)\\ &= (x+3)(x-5)(x+8) \left(\frac{1}{x+3} + \frac{1}{x-5} + \frac{1}{x+8}\right) \end{align*}$