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 $f(x) = \sqrt {x} \cos (x) \cdot e^x$ . Compute $\ddx f(x)\begin {prompt} = \answer {\frac {1}{2\sqrt {x}}\cos (x)e^x - \sqrt {x}\sin (x)e^x + \sqrt {x}\cos (x) e^x}\end {prompt}$

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

$\begin{align*} f(x) &= \sqrt{x}\cdot\cos(x)\cdot e^x\\ \ln(f(x)) &= \ln( \sqrt{x}\cdot\cos(x)\cdot e^x)\\ \ln(f(x)) &=\frac{1}{2}\ln{(x)} + \ln(\cos(x)) + \ln(e^x) \end{align*}$

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

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

Compute $\ddx \ln (e^x) \begin {prompt}=\answer {1}\end {prompt}$

So we have

$\begin{align*} \frac{f'(x)}{f(x)} &= \frac{1}{2x} - \frac{\sin(x)}{\cos(x)} + 1\\ f'(x) &= f(x) \left(\frac{1}{2x} - \frac{\sin(x)}{\cos(x)} + 1\right)\\ &= \sqrt{x}\cos(x)e^x\left(\frac{1}{2x} - \frac{\sin(x)}{\cos(x)} + 1\right) \end{align*}$