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Strategy in differentiating functions

AP.CALC:
FUN‑3 (EU)
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
Powerstart fraction, d, divided by, d, x, end fraction, open bracket, x, start superscript, n, end superscript, close bracket, equals, n, dot, x, start superscript, n, minus, 1, end superscript
Sumstart fraction, d, divided by, d, x, end fraction, open bracket, start color #11accd, f, left parenthesis, x, right parenthesis, end color #11accd, plus, start color #ca337c, g, left parenthesis, x, right parenthesis, end color #ca337c, close bracket, equals, start color #11accd, f, prime, left parenthesis, x, right parenthesis, end color #11accd, plus, start color #ca337c, g, prime, left parenthesis, x, right parenthesis, end color #ca337c
Productstart fraction, d, divided by, d, x, end fraction, open bracket, start color #11accd, f, left parenthesis, x, right parenthesis, end color #11accd, dot, start color #ca337c, g, left parenthesis, x, right parenthesis, end color #ca337c, close bracket, equals, start color #11accd, f, prime, left parenthesis, x, right parenthesis, end color #11accd, start color #ca337c, g, left parenthesis, x, right parenthesis, end color #ca337c, plus, start color #11accd, f, left parenthesis, x, right parenthesis, end color #11accd, start color #ca337c, g, prime, left parenthesis, x, right parenthesis, end color #ca337c
Quotientstart fraction, d, divided by, d, x, end fraction, open bracket, start fraction, start color #11accd, f, left parenthesis, x, right parenthesis, end color #11accd, divided by, start color #ca337c, g, left parenthesis, x, right parenthesis, end color #ca337c, end fraction, close bracket, equals, start fraction, start color #11accd, f, prime, left parenthesis, x, right parenthesis, end color #11accd, start color #ca337c, g, left parenthesis, x, right parenthesis, end color #ca337c, minus, start color #11accd, f, left parenthesis, x, right parenthesis, end color #11accd, start color #ca337c, g, prime, left parenthesis, x, right parenthesis, end color #ca337c, divided by, open bracket, start color #ca337c, g, left parenthesis, x, right parenthesis, end color #ca337c, close bracket, squared, end fraction
Chainstart fraction, d, divided by, d, x, end fraction, open bracket, start color #11accd, f, left parenthesis, start color #ca337c, g, left parenthesis, x, right parenthesis, end color #ca337c, right parenthesis, end color #11accd, close bracket, equals, start color #11accd, f, prime, left parenthesis, start color #ca337c, g, left parenthesis, x, right parenthesis, end color #ca337c, right parenthesis, end color #11accd, dot, start color #ca337c, g, prime, left parenthesis, x, right parenthesis, end color #ca337c
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 sine, left parenthesis, x, right parenthesis, 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 start color #11accd, left parenthesis, x, squared, plus, 1, right parenthesis, end color #11accd, dot, start color #ca337c, sine, left parenthesis, x, right parenthesis, end color #ca337c, 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 start fraction, start color #11accd, square root of, x, end square root, end color #11accd, divided by, start color #ca337c, cosine, left parenthesis, x, right parenthesis, end color #ca337c, end fraction, 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 parenthesis, 2, x, squared, minus, 4, right parenthesis, start superscript, 5, end superscript, try to think of it as an inner function and an outer function:
start color #11accd, start underbrace, left parenthesis, space, start color #ca337c, start overbrace, 2, x, squared, minus, 4, end overbrace, start superscript, start text, i, n, n, e, r, end text, end superscript, space, end color #ca337c, right parenthesis, start superscript, 5, end superscript, end underbrace, start subscript, start text, o, u, t, e, r, end text, end subscript, end color #11accd
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 parenthesis, x, squared, plus, 5, x, right parenthesis, dot, sine, left parenthesis, x, right parenthesis. Here is his work:
=ddx[(x2+5x)sin(x)]=ddx[x2+5x]ddx[sin(x)]=(2x+5)cos(x)=2xcos(x)+5cos(x)\begin{aligned} &\phantom{=}\dfrac{d}{dx}[(x^2+5x)\cdot\sin(x)] \\\\ &=\dfrac{d}{dx}[x^2+5x]\cdot\dfrac{d}{dx}[\sin(x)] \\\\ &=(2x+5)\cdot\cos(x) \\\\ &=2x\cdot\cos(x)+5\cdot\cos(x) \end{aligned}
Is Jake's work correct? If not, what's his mistake?
Choose 1 answer:
Choose 1 answer:

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 sine, left parenthesis, x, squared, plus, 5, x, right parenthesis. Here is his work:
=ddx[sin(x2+5x)]=ddx[sin(x)(x2+5x)]=ddx[sin(x)](x2+5x)+sin(x)ddx[x2+5x]=cos(x)(x2+5x)+sin(x)(2x+5)\begin{aligned} &\phantom{=}\dfrac{d}{dx}[\sin(x^2+5x)] \\\\ &=\dfrac{d}{dx}[\sin(x)\cdot(x^2+5x)] \\\\ &=\dfrac{d}{dx}[\sin(x)]\cdot(x^2+5x)+\sin(x)\cdot\dfrac{d}{dx}[x^2+5x] \\\\ &=\cos(x)(x^2+5x)+\sin(x)(2x+5) \end{aligned}
Is Leon's work correct? If not, what's his mistake?
Choose 1 answer:
Choose 1 answer:

Common mistake: Confusing function notation with multiplication

As we saw in Problem 2, start color #11accd, sine, left parenthesis, start color #ca337c, x, squared, plus, 5, end color #ca337c, right parenthesis, end color #11accd is a composite function where the outer function is start color #11accd, sine, left parenthesis, x, right parenthesis, end color #11accd and the inner function is start color #ca337c, x, squared, plus, 5, end color #ca337c. However, some people are confused by the notation and consider this to be the product start color #11accd, sine, left parenthesis, x, right parenthesis, end color #11accd, start color #ca337c, left parenthesis, x, squared, plus, 5, right parenthesis, end color #ca337c. 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 parenthesis, x, plus, 5, right parenthesis, left parenthesis, x, minus, 3, right parenthesis, but that would be a lot more work than what's needed. Instead, we can just expand the expression to x, squared, plus, 2, x, minus, 15 then apply the power rule to get the derivative: 2, x, plus, 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
=ddx[(x+5)(x3)]=ddx[x+5](x3)+(x+5)ddx[x3]=(1)(x3)+(x+5)(1)=x3+x+5=2x+2\begin{aligned}&\phantom{=}\dfrac{d}{dx}[(x+5)(x-3)]\\\\&=\dfrac{d}{dx}[x+5]\cdot (x-3)+(x+5)\cdot \dfrac{d}{dx}[x-3]\\\\&=(1)(x-3)+(x+5)(1)\\\\&=x-3+x+5\\\\&=2x+2\end{aligned}=ddx[(x+5)(x3)]=ddx[x2+2x15]=2x+2\begin{aligned}&\phantom{=}\dfrac{d}{dx}[(x+5)(x-3)]\\\\&=\dfrac{d}{dx}[x^2+2x-15]\\\\&=2x+2\end{aligned}
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 parenthesis, x, right parenthesis, equals, left parenthesis, 3, minus, 8, x, right parenthesis, left parenthesis, 2, x, minus, 7, right parenthesis
How would you rewrite f, left parenthesis, x, right parenthesis so it can be differentiated using the power rule?
Choose 1 answer:
Choose 1 answer:

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

We could apply the quotient rule to find the derivative of start fraction, x, start superscript, 6, end superscript, minus, 8, x, cubed, divided by, 2, x, squared, end fraction. However, it would be easier to divide first, getting 0, point, 5, x, start superscript, 4, end superscript, minus, 4, x, then apply the power rule to get the derivative of 2, x, cubed, minus, 4. We just have to remember that the function is undefined for x, equals, 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, start fraction, x, squared, plus, 5, x, minus, 14, divided by, x, minus, 7, end fraction 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 parenthesis, x, right parenthesis, equals, start fraction, x, start superscript, 5, end superscript, minus, 2, x, cubed, minus, 8, x, squared, divided by, x, end fraction
How would you rewrite f, left parenthesis, x, right parenthesis so it can be differentiated using the power rule?
Assume start text, x, end text, does not equal, start text, 0, end text.
Choose 1 answer:
Choose 1 answer:

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 start fraction, square root of, x, plus, 3, end square root, divided by, x, start superscript, 4, end superscript, end fraction 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.
x+3x4=x+31x4=x+3x4\begin{aligned}\dfrac{\sqrt{x+3}}{x^4}&=\sqrt{x+3}\cdot \dfrac{1}{x^4} \\\\ &=\sqrt{x+3}\cdot x^{-4} \end{aligned}
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 parenthesis, x, right parenthesis, equals, start fraction, sine, left parenthesis, x, right parenthesis, divided by, 3, x, end fraction
How would you rewrite h, left parenthesis, x, right parenthesis so it can be differentiated using the product rule?
Choose 1 answer:
Choose 1 answer:

Want more practice? Try this exercise.
A common struggle: It can be tricky to convert radicals or reciprocals into powers if you're uncomfortable with the process (examples: square root of, x, end square root, equals, x, start superscript, 1, slash, 2, end superscript and start fraction, 1, divided by, x, cubed, end fraction, equals, x, start superscript, minus, 3, end superscript). If you'd like some additional practice with this, check out these exercises:

Summary

Being fluent at taking derivatives requires knowing which rule to apply and when to apply it. It also requires seeing opportunities to rewrite expressions to make differentiation easier.
Here's a flowchart that summarizes this process:
A flowchart summarizes 2 steps, as follows. Step 1. Categorize the function. The 3 categories are product or quotient, composite, and basic function. Examples of basic functions include x to the n power, sine of x, cosine of x, e to the x power, and natural log of x. If function is a product or quotient, ask the question, can you change the function into another form that's easier to differentiate? If yes, revise the function to something that's easier to differentiate, then go back to step 1. If no, go to step 2. If the function is a composite or basic function, go to step 2. Step 2 is differentiate using the appropriate derivative rule.

Want to join the conversation?

  • primosaur ultimate style avatar for user Steve
    Great summary! But the only thing is..isn't this list missing the whole Limit approaching 0 method?
    (7 votes)
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    • female robot grace style avatar for user tyersome
      In general we only use limits when we want to prove something — e.g. to prove that these derivative rules are valid.

      Once the rules are proven for a general case there is no need go back to the formal definition of the derivative for specific problems.
      (22 votes)
  • blobby green style avatar for user Rubytranhcm
    why we need to consider the domain when cancelling the polynomial anyone pls help?
    (4 votes)
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    • starky ultimate style avatar for user KLaudano
      Suppose y(x) is a rational function defined to be (f(x)*h(x)) / (g(x)*h(x)) where f, g, and h are polynomial functions. When h(x) equals 0, f(x) is undefined. So if we cancel out h(x), we must be sure to exclude the values of x where h(x) = 0 from the domain of y(x).
      (3 votes)
  • male robot hal style avatar for user scott.d.corwin
    Speaking of calculation mistakes - I'm wondering how common they are. How often do you make mistakes when solving problems? To be clear, these are the type of mistakes that you make when you know how to do the work, but accidentally added 3 × 3, or your pencil wrote the opposite of what you were thinking. That kind of mistake.
    I counted my errors from the past 20 problems and missed 3, which feels about right. Because I always try to get 100% on everything prior to a unit test, I have spent lots of time going through a 4-problem exercise where I'd keep missing 1 problem then intentionally skipping the others to re-do. I recall spending the most time on trig word problems where you have to model a full function including phase shift.
    (4 votes)
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  • blobby green style avatar for user Pranathi
    what happens to the units of a function when it is differentiated?
    (2 votes)
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    • leafers ultimate style avatar for user Yan
      If we take the derivative of a function y=f(x), the unit becomes y unit/x unit.

      A derivative is the tangent line's slope, which is y/x. So the unit of the differentiated function will be the quotient.

      For example, v(t) is the derivative of s(t).
      s -> position -> unit: meter
      t -> time -> unit: second
      v (ds/dt, the derivative) -> speed -> unit: meter per second
      (2 votes)
  • leafers seed style avatar for user Hongbing Zhu
    I am confused if we can rewrite the square root of (3 times (cos(x))'s third power) to 3's square root times cos(x)'s 3/2 power, then appply power rule? it is seems different result compared with if firstly apply chain rule to the whole? this is one question in the last "practice", thanks.
    (1 vote)
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    • cacteye blue style avatar for user Jerry Nilsson
      𝑑∕𝑑𝑥√(3 cos³𝑥) =
      = 1∕(2√(3 cos³𝑥)) ∙ 3 ∙ 3 cos²𝑥 ∙ (−sin 𝑥) =
      = −3√3√3 cos²𝑥 sin 𝑥∕(2√3 cos 𝑥 √(cos 𝑥)) =
      = −(3√3√(cos 𝑥) sin 𝑥)∕2

      𝑑∕𝑑𝑥(√3 ∙ (cos 𝑥)^(3∕2) = √3 ∙ (3∕2) (cos 𝑥)^(1∕2) ∙ (−sin 𝑥) =
      = −(3√3√(cos 𝑥) sin 𝑥)∕2
      (1 vote)
  • leaf green style avatar for user Yu Aoi
    are there other rules? this is not all?
    (0 votes)
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