🌌 Calculus Kingdom · Calculus
Derivative Rules
Differentiate products, quotients and compositions with the product, quotient and chain rules.
In short
- Product rule: (uv)' = u'v + uv' — the derivative of a product is not the product of the derivatives.
- Quotient rule: (u/v)' = (u'v - uv')/v2, with the u' term first and the denominator squared.
- Chain rule: differentiate the outside, keep the inside, then multiply by the derivative of the inside.
- Start from the outermost operation, and expect to combine rules on harder functions.
Products are not as simple as sums
Sums differentiate term by term, and it is tempting to assume products behave the same way. They do not.
Test it: for f(x) = x x x2 = x3 the derivative is 3x2, but multiplying the separate derivatives would give 1 x 2x = 2x. Not the same.
The correct rule is the product rule:
(u v)' = u' v + u v'
Differentiate the first and keep the second, then keep the first and differentiate the second, and add the two pieces.
For f(x) = (2x + 3)(x - 5): here u = 2x + 3 with u' = 2, and v = x - 5 with v' = 1, so
f'(x) = 2(x - 5) + 1(2x + 3) = 4x - 7
Checking by expanding first gives 2x2 - 7x - 15, whose derivative is 4x - 7. They agree.
The quotient rule
For a fraction of two functions:
(u / v)' = (u' v - u v') / v2
Three details matter and all three get lost regularly:
- the order — the u' term comes first, and the middle sign is a minus, so the rule is not symmetric
- the denominator is squared
- the whole numerator sits over that square
For f(x) = (3x + 1)/(x - 2):
u' v - u v' = 3(x - 2) - 1(3x + 1) = 3x - 6 - 3x - 1 = -7 f'(x) = -7 / (x - 2)2
The x-terms cancelling like that is typical when both parts are linear, and it makes a good check on your algebra.
Where a quotient can be rewritten as a product or a power, that is usually easier. 5/x3 is much less trouble as 5x-3.
The chain rule
The chain rule handles a function inside another function:
(f(g(x)))' = f'(g(x)) x g'(x)
In words: differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
d/dx (3x + 1)5 = 5(3x + 1)4 x 3 = 15(3x + 1)4
That final factor of 3 is the whole point, and forgetting it is the single most common error in all of calculus. It has a name — forgetting the chain rule — and it is worth checking for every time.
Why the extra factor? If the inside changes three times as fast as x, then the outside is being fed a rapidly changing input, and its own rate of change is multiplied accordingly.
The same pattern covers the standard functions:
d/dx sqrt(3x + 1) = 3 / (2 sqrt(3x + 1)) d/dx e4x = 4 e4x d/dx sin(5x) = 5 cos(5x) d/dx cos(5x) = -5 sin(5x)
Trigonometric derivatives hold only when x is in radians, and the minus sign on the derivative of cosine is part of the rule.
Choosing and combining rules
Look at the outermost operation to decide where to start.
- two things multiplied — product rule
- one thing divided by another — quotient rule
- one function wrapped inside another (a bracket to a power, a root, an exponential, a trigonometric function of an expression) — chain rule
Real problems combine them. For f(x) = x(2x + 1)4 the outermost structure is a product, so start there with u = x and v = (2x + 1)4. Finding v' then needs the chain rule:
v' = 4(2x + 1)3 x 2 = 8(2x + 1)3 f'(x) = (2x + 1)4 + 8x(2x + 1)3
Work from the outside in, and write each piece down before assembling. Trying to do it all in one line is where mistakes breed.
If you are unsure whether an answer is right, test it numerically: compute the average slope of f over a tiny interval around some x, and compare with your derivative there. They should nearly match.
Worked examples
Example 1
f(x) = (x2 + 1)(3x - 2). Find f'(x).
- This is a product, so use the product rule with u = x2 + 1 and v = 3x - 2.
- u' = 2x and v' = 3.
- f'(x) = u'v + uv' = 2x(3x - 2) + 3(x2 + 1).
- Expanding gives 6x2 - 4x + 3x2 + 3 = 9x2 - 4x + 3.
Example 2
f(x) = (4x - 7)6. Find f'(x).
- One expression sits inside a power, so this is the chain rule.
- Outer: something^6 differentiates to 6 x something^5, with the inside untouched: 6(4x - 7)5.
- Inner: 4x - 7 has derivative 4, so multiply by 4.
- f'(x) = 24(4x - 7)5.
Practice problems, with solutions
Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.
Problem 1
Difficulty 1 of 5f(x) = (3x + 2)(3x + 3) Find f'(5).
Answer: 105
- u = 3x + 2, u' = 3; v = 3x + 3, v' = 3
- f'(x) = u'v + uv' = 3(3x + 3) + 3(3x + 2)
- f'(5) = 3(18) + 3(17)
- f'(5) = 105
Problem 2
Difficulty 3 of 5f(x) = (5x + 8) / (3x + 5) Find f'(x). Give your answer as an expression in x.
Answer: 1/(3x + 5)2
- f'(x) = (u'v - uv') / v2
- u'v - uv' = 5(3x + 5) - 3(5x + 8)
- = 15x + 25 - 24 - 15x = 1
- f'(x) = 1 / (3x + 5)2
Problem 3
Difficulty 4 of 5f(x) = (-5x + 1)2 Find f'(x). Give your answer as an expression in x (it need not be expanded).
Answer: -10(-5x + 1)
- Outer: u2 -> 2u1, with u = -5x + 1
- Inner: u = -5x + 1 -> u' = -5
- f'(x) = 2(-5x + 1)1 x -5
- f'(x) = -10(-5x + 1)1
Common mistakes
- Forgetting the chain rule, so the inner derivative is never multiplied in.
- Multiplying the two derivatives together instead of using the product rule.
- Reversing the numerator of the quotient rule, or forgetting to square the denominator.
- Applying the power rule to ex, whose variable is in the exponent rather than the base.
What you should be able to do
- Apply the product rule.
- Apply the quotient rule.
- Apply the chain rule to a composition.
- Combine rules on a single expression.
Where this fits in the curriculum
Common Core
- FUN-3.B
AP Calculus AB, Unit 2 — Calculate derivatives of products and quotients of differentiable functions.
The Common Core has no calculus standards; this is the AP Calculus AB learning objective.
- FUN-3.C
AP Calculus AB, Unit 3 — Calculate derivatives of compositions of differentiable functions (the chain rule).
The Common Core has no calculus standards; this is the AP Calculus AB learning objective.