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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 3

Lesson 6: Selecting procedures for calculating derivatives: strategy

# Strategy in differentiating functions

Differentiation has so many different rules and there are so many different ways to apply them! Let's take a broader look at differentiation and come up with a workflow that will allow us to find the derivative of any function, efficiently and without mistakes.
Many calculus students know their derivative rules pretty well yet struggle to apply the right rule in the right situation. To alleviate this struggle, we want to learn to quickly categorize functions, know which rule to apply, and even rewrite functions in different forms to make differentiation easier.
For reference, here is a summary of the most common derivative rules:
NameRule
Power$\frac{d}{dx}\left[{x}^{n}\right]=n\cdot {x}^{n-1}$
Sum$\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]={f}^{\prime }\left(x\right)+{g}^{\prime }\left(x\right)$
Product$\frac{d}{dx}\left[f\left(x\right)\cdot g\left(x\right)\right]={f}^{\prime }\left(x\right)g\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right)$
Quotient$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{{f}^{\prime }\left(x\right)g\left(x\right)-f\left(x\right){g}^{\prime }\left(x\right)}{\left[g\left(x\right){\right]}^{2}}$
Chain$\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]={f}^{\prime }\left(g\left(x\right)\right)\cdot {g}^{\prime }\left(x\right)$
We'll be focusing on the last three rules because those are usually the most challenging to apply.

## Spotting products, quotients, and compositions

Most derivative rules tell us how to differentiate a specific kind of function, like the rule for the derivative of $\mathrm{sin}\left(x\right)$, or the power rule.
However, there are three very important rules that are generally applicable, and depend on the structure of the function we are differentiating. These are the product, quotient, and chain rules, so be on the lookout for them. Ask yourself: "Do I see a product, quotient, or composition of functions?"
Product: If you see something like $\left({x}^{2}+1\right)\cdot \mathrm{sin}\left(x\right)$, you want to notice that this is the product of two functions. From there, you can apply the product rule.
Quotient: Similarly, if you see something like $\frac{\sqrt{x}}{\mathrm{cos}\left(x\right)}$, you want to notice that one function is being divided by the other, and that the quotient rule will apply.
Composition: Lastly, if you see a function like $\left(2{x}^{2}-4{\right)}^{5}$, try to think of it as an inner function and an outer function:
This sort of function is called a composite function, and you can apply the chain rule to find its derivative.
Problem 1
Jake tried to find the derivative of $\left({x}^{2}+5x\right)\cdot \mathrm{sin}\left(x\right)$. Here is his work:
$\begin{array}{rl}& \phantom{=}\frac{d}{dx}\left[\left({x}^{2}+5x\right)\cdot \mathrm{sin}\left(x\right)\right]\\ \\ & =\frac{d}{dx}\left[{x}^{2}+5x\right]\cdot \frac{d}{dx}\left[\mathrm{sin}\left(x\right)\right]\\ \\ & =\left(2x+5\right)\cdot \mathrm{cos}\left(x\right)\\ \\ & =2x\cdot \mathrm{cos}\left(x\right)+5\cdot \mathrm{cos}\left(x\right)\end{array}$
Is Jake's work correct? If not, what's his mistake?

### Common mistake: forgetting to apply the product or quotient rules

Remember: Taking the product of the derivatives is not the same as applying the product rule.
Similarly, taking the quotient of the derivatives is not the same as applying the quotient rule.
Problem 2
Leon tried to find the derivative of $\mathrm{sin}\left({x}^{2}+5x\right)$. Here is his work:
$\begin{array}{rl}& \phantom{=}\frac{d}{dx}\left[\mathrm{sin}\left({x}^{2}+5x\right)\right]\\ \\ & =\frac{d}{dx}\left[\mathrm{sin}\left(x\right)\cdot \left({x}^{2}+5x\right)\right]\\ \\ & =\frac{d}{dx}\left[\mathrm{sin}\left(x\right)\right]\cdot \left({x}^{2}+5x\right)+\mathrm{sin}\left(x\right)\cdot \frac{d}{dx}\left[{x}^{2}+5x\right]\\ \\ & =\mathrm{cos}\left(x\right)\left({x}^{2}+5x\right)+\mathrm{sin}\left(x\right)\left(2x+5\right)\end{array}$
Is Leon's work correct? If not, what's his mistake?

### Common mistake: Confusing function notation with multiplication

As we saw in Problem 2, $\mathrm{sin}\left({x}^{2}+5\right)$ is a composite function where the outer function is $\mathrm{sin}\left(x\right)$ and the inner function is ${x}^{2}+5$. However, some people are confused by the notation and consider this to be the product $\mathrm{sin}\left(x\right)\left({x}^{2}+5\right)$. This is an entirely different function, and differentiating it will result in a wrong derivative.

## We can rewrite functions to make differentiation easier.

Let's face it: applying the product, quotient, and chain rules can be a lot of work. The quotient rule is especially demanding. So why would we do all of that work if we don't have to? The following three examples highlight some products and quotients that can be rewritten to make differentiation much easier.
Making expressions more efficient to differentiate isn't just a matter of convenience; The simpler and shorter the differentiation, the smaller the chance that you make a mistake along the way!

### Sometimes, we can rewrite a product as a simple polynomial.

We could apply the product rule to differentiate $\left(x+5\right)\left(x-3\right)$, but that would be a lot more work than what's needed. Instead, we can just expand the expression to ${x}^{2}+2x-15$ then apply the power rule to get the derivative: $2x+2$.
To really drive home the point, just look at how much more work it would have been to use the product rule:
Product rulePower rule
$\begin{array}{rl}& \phantom{=}\frac{d}{dx}\left[\left(x+5\right)\left(x-3\right)\right]\\ \\ & =\frac{d}{dx}\left[x+5\right]\cdot \left(x-3\right)+\left(x+5\right)\cdot \frac{d}{dx}\left[x-3\right]\\ \\ & =\left(1\right)\left(x-3\right)+\left(x+5\right)\left(1\right)\\ \\ & =x-3+x+5\\ \\ & =2x+2\end{array}$$\begin{array}{rl}& \phantom{=}\frac{d}{dx}\left[\left(x+5\right)\left(x-3\right)\right]\\ \\ & =\frac{d}{dx}\left[{x}^{2}+2x-15\right]\\ \\ & =2x+2\end{array}$
To be clear: both ways are correct, but using the power rule will take you less time, and has better chances of avoiding calculation mistakes along the way.
Problem 3
$f\left(x\right)=\left(3-8x\right)\left(2x-7\right)$
How would you rewrite $f\left(x\right)$ so it can be differentiated using the power rule?

### Similarly, some quotient rule problems can be rewritten to use the power rule

We could apply the quotient rule to find the derivative of $\frac{{x}^{6}-8{x}^{3}}{2{x}^{2}}$. However, it would be easier to divide first, getting $0.5{x}^{4}-4x$, then apply the power rule to get the derivative of $2{x}^{3}-4$. We just have to remember that the function is undefined for $x=0$, and therefore so is the derviative.
If we do it the long way, with the quotient rule, we get the same result. However, we have a better chance to make some kind of mistake along the way.
Not every quotient can be rewritten this way. For example, $\frac{{x}^{2}+5x-14}{x-7}$ cannot be simplified as a polynomial.
Remember: You can always use this method for quotients where the denominator is a monomial.
When the denominator is a polynomial with more than one term, you might be able to simplify it using factorization and canceling common terms.
Don't forget to consider the domain when rewriting quotients.
Problem 4
$f\left(x\right)=\frac{{x}^{5}-2{x}^{3}-8{x}^{2}}{x}$
How would you rewrite $f\left(x\right)$ so it can be differentiated using the power rule?
Assume $\mathit{\text{x}}\ne \mathit{\text{0}}$.

### Last example: rewriting a quotient as a product

For many people, the product rule is easier to remember than the quotient rule is. Fortunately, we can always rewrite a quotient as a product.
Suppose we wanted to differentiate $\frac{\sqrt{x+3}}{{x}^{4}}$ but couldn't remember the order of the terms in the quotient rule. We could first separate the numerator and denominator into separate factors, then rewrite the denominator using a negative exponent so we would have no quotients.
$\begin{array}{rl}\frac{\sqrt{x+3}}{{x}^{4}}& =\sqrt{x+3}\cdot \frac{1}{{x}^{4}}\\ \\ & =\sqrt{x+3}\cdot {x}^{-4}\end{array}$
Now we would be ready to use our product rule. (Note: we would also use the chain rule to handle the interior of the square root function.)
Problem 5
$h\left(x\right)=\frac{\mathrm{sin}\left(x\right)}{3x}$
How would you rewrite $h\left(x\right)$ so it can be differentiated using the product rule?
A common struggle: It can be tricky to convert radicals or reciprocals into powers if you're uncomfortable with the process (examples: $\sqrt{x}={x}^{1/2}$ and $\frac{1}{{x}^{3}}={x}^{-3}$). If you'd like some additional practice with this, check out these exercises: