Question 1

When do I use the product rule vs. the chain rule?

Accepted Answer

Use the product rule when you are differentiating a function that is one function MULTIPLIED by another, like $f(x) = x^2 e^x$. Use the chain rule when you have a function INSIDE another function (a composite function), like $f(x) = e^{(x^2)}$. Sometimes you need both, for example in $f(x) = x^2 e^{3x}$ you use the product rule overall, but need the chain rule to differentiate the $e^{3x}$ part.

Question 2

Is there an easier way than the quotient rule?

Accepted Answer

Sometimes, yes. You can often rewrite a quotient as a product and use the product rule. For example, $y = \frac{\ln x}{x^2}$ can be written as $y = (\ln x)(x^{-2})$. Differentiating this with the product rule can sometimes feel simpler and lead to fewer sign errors. However, you must be comfortable with both methods.

Question 3

What's the point of the second derivative test if it can be inconclusive?

Accepted Answer

The second derivative test is usually much faster than testing points on either side of the stationary point. It's only inconclusive in specific cases (like $y=x^3$ or $y=x^4$ at $x=0$). For most exam questions, it works perfectly. You just need to remember the backup method (testing the sign of $f'(x)$) for when $f''(x)=0$.

Question 4

In connected rates of change, how do I know which variable to eliminate?

Accepted Answer

Look at the rate you are given and the rate you need to find. For example, if you are given $\frac{dV}{dt}$ and need to find $\frac{dh}{dt}$, your chain rule will be $\frac{dh}{dt} = \frac{dh}{dV} 	imes \frac{dV}{dt}$. This tells you that you need a relationship between $h$ and $V$. Therefore, you should write your volume formula $V$ in terms of $h$ only, eliminating other variables like the radius $r$.

Differentiation

What you need to know

The Calculus of Change