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*}