## Using this document

I recommend you think for 1-3 minutes after getting stuck before viewing a hint, and 10-15 minutes before viewing giving up and viewing the solution.

Pay attention to the lower *and upper* bounds! There's no honor in staring blankly at a problem for 30 minutes when you're stuck, and likewise viewing solutions prematurely is inefficient.

## Exercises

### Inverse and Implicit Differentiation

Using the fact that $(e^x)' = e^x$ and $e^{\ln x} = x$ find $(\ln x)'$

## Hint

Differentiate both sides of $e^{\ln x} = x$ then solve for $(\ln x)'$

## Solution

We know

Taking the derivative of both sides and using the chain rule gives

Since $e^{\ln x} = x$ dividing by $x$ isolates $(\ln x)'$ to give

### Logarithmic Differentiation

Derive the product rule $(fg)' = f'g+g'f$ using the chain rule and logarithms.

## Hint

Logarithms turn multiplication into addition, and we know how to handle addition (derivative of sum is sum of derivatives!)

## Solution

Let $h = fg$ (or more verbosely $h(x) = f(x)g(x)$) then take the log of both sides

Differentiate both sides, the chain rule and the fact that $(\ln x)' = 1/x$ imply $(\ln f(x))' = f'(x)/f(x)$ giving

Since $h = fg$ this becomes

Solving for $(fg)'$ by multiplying both sides by $fg$ finally gives the familiar product rule

Generalize the product rule to a product $f_1\dots f_n$ of more than two functions.

## Hint

Use the logarithm approach from the previous exercise.

## Solution

Let $h = f_1\dots f_n$, take logs to get

Take derivatives

Multiply both sides by $h$ and substitute back in for $h$

Compute the following derivative

## Hint

Write

and take the logarithm of both sides

## Solution

Take the log of both sides and simplify

Take derivatives to get

Multiply by $y$ and substitute back in

### Deriving Identities

In probability, we need to compute the following sum to find the expected value of a random variable with a geometric distribution

Using the identity $\sum_{k=0}^\infty q^k = \frac{1}{1-q}$ find the sum above

## Hint

Differentiate both sides of the identity with respect to $q$

## Solution

Multiply both sides by $q$ to turn $q^{k-1}$ into $q^k$, multiply by $p$ to get the desired sum

(Footnote about probability: since $q=1-p$ the expectation simplifies to $1/p - 1$.)