Differentiation rules

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1. The power, constant multiple and sum rules

Warm-up

Three quick questions from earlier lessons.

Discover

From the limit definition: (x2)′=2x(x^2)' = 2x and (x3)′=3x2(x^3)' = 3x^2. Guess the pattern: what is (x4)′(x^4)'?

  1. 4x34x^3
  2. x3x^3

Key idea

Three rules replace the limit for every polynomial and power.

Power rule: (xn)′=n xn−1(x^n)' = n\,x^{n-1} for every real nn. Constant multiple: (k f)′=k f′(k\,f)' = k\,f'. Sum: (f+g)′=f′+g′(f + g)' = f' + g'. Constant: (c)′=0(c)' = 0.

Worked example

f(x)=5x3−2x+7f(x) = 5x^3 - \dfrac{2}{x} + 7. Find f′(x)f'(x).

f(x)=5x3−2x−1+7f(x) = 5x^3 - 2x^{-1} + 7
Write the fraction as a power.
f′(x)=15x2+2x−2f'(x) = 15x^2 + 2x^{-2}
Power rule term by term; the constant 7 gives 0.
f′(x)=15x2+2x2f'(x) = 15x^2 + \dfrac{2}{x^2}
Back to a fraction if you like.

Your turn, guided

Your turn: f(x)=4x5+3x−9f(x) = 4x^5 + 3x - 9.

f′(x)=20x4+3f'(x) = 20x^4 + 3
20x⁴ from 4x⁵, 3 from 3x, 0 from −9.

Practice ★

  1. f(x)=8x5+3xf(x) = 8x^{ 5 } + 3x. Find f′(x)f'(x).
  2. f(x)=4x3+6xf(x) = 4x^{ 3 } + 6x. Find f′(x)f'(x).
  3. f(x)=6x5+5xf(x) = 6x^{ 5 } + 5x. Find f′(x)f'(x).

Watch out!

Practice ★★

  1. f(x)=6x2f(x) = \dfrac{ 6 }{x^{ 2 } }. Find f′(x)f'(x).
  2. f(x)=9x3f(x) = \dfrac{ 9 }{x^{ 3 } }. Find f′(x)f'(x).
  1. f(x)=3x+8f(x) = 3\sqrt{x} + 8. Find f′(x)f'(x).
  2. f(x)=5x+4f(x) = 5\sqrt{x} + 4. Find f′(x)f'(x).

Exit ticket

Three questions to check you have got it.

  1. f(x)=8x5+3xf(x) = 8x^{ 5 } + 3x. Find f′(x)f'(x).
  2. f(x)=4x3+6xf(x) = 4x^{ 3 } + 6x. Find f′(x)f'(x).
  3. f(x)=6x5+5xf(x) = 6x^{ 5 } + 5x. Find f′(x)f'(x).

Summary card

Basic rules

  • (xn)′=nxn−1(x^n)' = n x^{n-1}; rewrite roots and fractions as powers first.
  • Constants factor out; sums split; a constant alone gives 0.

Watch out: with negative exponents, −2−1=−3-2 - 1 = -3.

2. Derivatives of exponential functions

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is y=exy = e^x, the blue line its tangent at x=ax = a. Move aa and compare the slope of the line with the height of the point.

The curve y = eˣ and its tangent at x = a. The tangent's slope equals the height eᵃ of the point of contact.

At a=0a = 0 the point is (0,1)(0, 1). What is the slope of the tangent there?

  1. 1
  2. 0

Key idea

The exponent varies, so the power rule does not apply.

(ex)′=ex(e^x)' = e^x. For any base b>0b > 0: (bx)′=bxln⁡b(b^x)' = b^x \ln b.

Worked example

f(x)=3ex−2xf(x) = 3e^x - 2^x. Find f′(x)f'(x).

(3ex)′=3ex(3e^x)' = 3e^x
eˣ is its own derivative.
(2x)′=2xln⁡2(2^x)' = 2^x \ln 2
Base 2: multiply by ln 2.
f′(x)=3ex−2xln⁡2f'(x) = 3e^x - 2^x \ln 2
Combine.

Your turn, guided

Your turn: f(x)=5ex+x3f(x) = 5e^x + x^3.

f′(x)=5ex+3x2f'(x) = 5e^x + 3x^2
5eˣ stays; x³ gives 3x².

Practice ★

  1. f(x)=8ex+5x2f(x) = 8e^x + 5x^2. Find f′(x)f'(x).
  2. f(x)=4ex+6x2f(x) = 4e^x + 6x^2. Find f′(x)f'(x).
  3. f(x)=7ex+4x2f(x) = 7e^x + 4x^2. Find f′(x)f'(x).

Watch out!

Practice ★★

  1. f(x)=3xf(x) = 3^x. Find f′(x)f'(x).
  2. f(x)=4xf(x) = 4^x. Find f′(x)f'(x).
  1. Find the slope of the tangent line to y=8ex+1y = 8e^x + 1 at x=0x = 0.
  2. Find the slope of the tangent line to y=2ex+4y = 2e^x + 4 at x=0x = 0.

Exit ticket

Three questions to check you have got it.

  1. f(x)=3xf(x) = 3^x. Find f′(x)f'(x).
  2. f(x)=4xf(x) = 4^x. Find f′(x)f'(x).
  3. f(x)=2xf(x) = 2^x. Find f′(x)f'(x).

Summary card

Exponentials

  • (ex)′=ex(e^x)' = e^x.
  • (bx)′=bxln⁡b(b^x)' = b^x \ln b.

Watch out: xnx^n (variable base) and bxb^x (variable exponent) have different rules.

3. Derivatives of sine and cosine

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curves are y=sin⁡xy = \sin x and y=cos⁡xy = \cos x; the blue line is the tangent to sin⁡x\sin x at x=ax = a. Move aa and compare the slope of the line with the height of the cosine curve at x=ax = a.

The sine curve, the cosine curve and the tangent to the sine curve at x = a. The tangent's slope is always the height of the cosine curve at a.

At a=π/2≈1.57a = \pi/2 \approx 1.57, sine is at its top. What is the slope there, and what is cos⁡(π/2)\cos(\pi/2)?

  1. Both are 0
  2. Both are 1

Key idea

With xx in radians:

(sin⁡x)′=cos⁡x(\sin x)' = \cos x and (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x.

Worked example

f(x)=3sin⁡x−4cos⁡xf(x) = 3\sin x - 4\cos x. Find f′(x)f'(x).

(3sin⁡x)′=3cos⁡x(3\sin x)' = 3\cos x
(sin x)' = cos x.
(−4cos⁡x)′=4sin⁡x(-4\cos x)' = 4\sin x
(cos x)' = −sin x, and − · − = +.
f′(x)=3cos⁡x+4sin⁡xf'(x) = 3\cos x + 4\sin x
Combine.

Your turn, guided

Your turn: f(x)=2cos⁡x+x2f(x) = 2\cos x + x^2.

f′(x)=−2sin⁡x+2xf'(x) = -2\sin x + 2x
−2 sin x and 2x.

Practice ★

  1. f(x)=3sin⁡x+5cos⁡xf(x) = 3\sin x + 5\cos x. Find f′(x)f'(x).
  2. f(x)=5sin⁡x+7cos⁡xf(x) = 5\sin x + 7\cos x. Find f′(x)f'(x).
  3. f(x)=5sin⁡x+9cos⁡xf(x) = 5\sin x + 9\cos x. Find f′(x)f'(x).

Watch out!

Practice ★★

  1. Find the slope of the tangent line to y=sin⁡xy = \sin x at x=π2x = \dfrac{\pi}{ 2 }. Give an exact value.
  2. Find the slope of the tangent line to y=sin⁡xy = \sin x at x=π3x = \dfrac{\pi}{ 3 }. Give an exact value.
  1. f(x)=4x−4cos⁡xf(x) = 4x - 4\cos x. Find f′(x)f'(x).
  2. f(x)=4x−8cos⁡xf(x) = 4x - 8\cos x. Find f′(x)f'(x).

Exit ticket

Three questions to check you have got it.

  1. f(x)=3sin⁡x+5cos⁡xf(x) = 3\sin x + 5\cos x. Find f′(x)f'(x).
  2. f(x)=5sin⁡x+7cos⁡xf(x) = 5\sin x + 7\cos x. Find f′(x)f'(x).
  3. f(x)=5sin⁡x+9cos⁡xf(x) = 5\sin x + 9\cos x. Find f′(x)f'(x).

Summary card

Sine and cosine

  • (sin⁡x)′=cos⁡x(\sin x)' = \cos x.
  • (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x.

Watch out: the minus sign goes with the derivative of cosine; radians only.

4. The product rule

Warm-up

Three quick questions from earlier lessons.

Discover

f(x)=xf(x) = x and g(x)=xg(x) = x, so f(x)g(x)=x2f(x)g(x) = x^2. Is (x2)′(x^2)' equal to f′(x) g′(x)=1⋅1f'(x)\,g'(x) = 1 \cdot 1?

  1. No
  2. Yes

Key idea

Each factor takes its turn to be differentiated.

Product rule: (f g)′=f′ g+f g′(f\,g)' = f'\,g + f\,g'.

Worked example

h(x)=x3exh(x) = x^3 e^x. Find h′(x)h'(x).

f′=3x2f' = 3x^2, g′=exg' = e^x
f = x³, g = eˣ.
h′(x)=3x2ex+x3exh'(x) = 3x^2 e^x + x^3 e^x
f'g + fg'.
h′(x)=x2ex(3+x)h'(x) = x^2 e^x (3 + x)
Factor if you like.

Your turn, guided

Your turn: h(x)=xcos⁡xh(x) = x\cos x.

f′=1f' = 1, g′=−sin⁡xg' = -\sin x
f = x, g = cos x.
h′(x)=cos⁡x−xsin⁡xh'(x) = \cos x - x\sin x
f'g + fg'.

Practice ★

  1. f(x)=x2sin⁡xf(x) = x^{ 2 }\sin x. Find f′(x)f'(x).
  2. f(x)=x3sin⁡xf(x) = x^{ 3 }\sin x. Find f′(x)f'(x).
  1. f(x)=(x+3)exf(x) = (x + 3)e^x. Find f′(x)f'(x).
  2. f(x)=(x+8)exf(x) = (x + 8)e^x. Find f′(x)f'(x).

Watch out!

Practice ★★

  1. f(2)=4f(2) = 4, f′(2)=4f'(2) = 4, g(2)=−4g(2) = -4 and g′(2)=−1g'(2) = -1. Find (fg)′(2)(fg)'(2).
  2. f(2)=−4f(2) = -4, f′(2)=−2f'(2) = -2, g(2)=2g(2) = 2 and g′(2)=3g'(2) = 3. Find (fg)′(2)(fg)'(2).
  3. f(2)=−1f(2) = -1, f′(2)=1f'(2) = 1, g(2)=5g(2) = 5 and g′(2)=−1g'(2) = -1. Find (fg)′(2)(fg)'(2).

Exit ticket

Three questions to check you have got it.

  1. f(x)=x2sin⁡xf(x) = x^{ 2 }\sin x. Find f′(x)f'(x).
  2. f(x)=x3sin⁡xf(x) = x^{ 3 }\sin x. Find f′(x)f'(x).
  3. f(x)=x5sin⁡xf(x) = x^{ 5 }\sin x. Find f′(x)f'(x).

Summary card

Product rule

(fg)′=f′g+fg′(fg)' = f'g + fg': differentiate one factor at a time, keep the other.

Watch out: (fg)′≠f′g′(fg)' \ne f'g'.

5. The quotient rule

Warm-up

Three quick questions from earlier lessons.

Discover

x3x=x2\dfrac{x^3}{x} = x^2, whose derivative is 2x2x. The quotient of the derivatives would be 3x21\dfrac{3x^2}{1}. Do they agree?

  1. No
  2. Yes

Key idea

Top's derivative first, then subtract.

Quotient rule: (fg)′=f′ g−f g′g2\left(\dfrac{f}{g}\right)' = \dfrac{f'\,g - f\,g'}{g^2}, where g≠0g \ne 0.

Worked example

h(x)=x2x+1h(x) = \dfrac{x^2}{x + 1}. Find h′(x)h'(x).

f′=2xf' = 2x, g′=1g' = 1
f = x², g = x + 1.
h′(x)=2x(x+1)−x2(x+1)2h'(x) = \dfrac{2x(x + 1) - x^2}{(x + 1)^2}
f'g − fg' over g².
h′(x)=x2+2x(x+1)2h'(x) = \dfrac{x^2 + 2x}{(x + 1)^2}
Simplify the top.

Your turn, guided

Your turn: h(x)=exxh(x) = \dfrac{e^x}{x}.

h′(x)=xex−exx2h'(x) = \dfrac{xe^x - e^x}{x^2}
f = eˣ, g = x: f'g − fg' = xeˣ − eˣ.

Practice ★

  1. f(x)=xx+6f(x) = \dfrac{x}{x + 6}. Find f′(x)f'(x).
  2. f(x)=xx+3f(x) = \dfrac{x}{x + 3}. Find f′(x)f'(x).
  3. f(x)=xx+2f(x) = \dfrac{x}{x + 2}. Find f′(x)f'(x).

Watch out!

Practice ★★

  1. f(x)=4sin⁡xxf(x) = \dfrac{ 4\sin x}{x}. Find f′(x)f'(x).
  2. f(x)=1sin⁡xxf(x) = \dfrac{ 1\sin x}{x}. Find f′(x)f'(x).
  1. f(1)=1f(1) = 1, f′(1)=−2f'(1) = -2, g(1)=4g(1) = 4 and g′(1)=2g'(1) = 2. Find (fg)′(1)\left(\dfrac{f}{g}\right)'(1).
  2. f(1)=1f(1) = 1, f′(1)=1f'(1) = 1, g(1)=1g(1) = 1 and g′(1)=4g'(1) = 4. Find (fg)′(1)\left(\dfrac{f}{g}\right)'(1).

Exit ticket

Three questions to check you have got it.

  1. f(x)=xx+6f(x) = \dfrac{x}{x + 6}. Find f′(x)f'(x).
  2. f(x)=xx+3f(x) = \dfrac{x}{x + 3}. Find f′(x)f'(x).
  3. f(x)=xx+2f(x) = \dfrac{x}{x + 2}. Find f′(x)f'(x).

Summary card

Quotient rule

(fg)′=f′g−fg′g2\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}: "low d-high minus high d-low, over low squared".

Watch out: the order in the numerator matters.

6. Derivatives of tangent, secant, cotangent and cosecant

Warm-up

Three quick questions from earlier lessons.

Discover

tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}. With the quotient rule, the top of (tan⁡x)′(\tan x)' is cos⁡x⋅cos⁡x−sin⁡x⋅(−sin⁡x)\cos x \cdot \cos x - \sin x \cdot (-\sin x). What does it simplify to?

  1. 1
  2. cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x

Key idea

The quotient rule gives the other four.

(tan⁡x)′=sec⁡2x(\tan x)' = \sec^2 x, (sec⁡x)′=sec⁡xtan⁡x(\sec x)' = \sec x \tan x, (cot⁡x)′=−csc⁡2x(\cot x)' = -\csc^2 x, (csc⁡x)′=−csc⁡xcot⁡x(\csc x)' = -\csc x \cot x. The "co" functions get a minus sign.

Worked example

f(x)=2tan⁡x−csc⁡xf(x) = 2\tan x - \csc x. Find f′(x)f'(x).

(2tan⁡x)′=2sec⁡2x(2\tan x)' = 2\sec^2 x
(tan x)' = sec² x.
(−csc⁡x)′=csc⁡xcot⁡x(-\csc x)' = \csc x \cot x
(csc x)' = −csc x cot x, and − · − = +.
f′(x)=2sec⁡2x+csc⁡xcot⁡xf'(x) = 2\sec^2 x + \csc x\cot x
Combine.

Your turn, guided

Your turn: f(x)=sec⁡x+xf(x) = \sec x + x.

f′(x)=sec⁡xtan⁡x+1f'(x) = \sec x\tan x + 1
sec x tan x, plus 1.

Practice ★

  1. f(x)=7tan⁡xf(x) = 7\tan x. Find f′(x)f'(x).
  2. f(x)=3tan⁡xf(x) = 3\tan x. Find f′(x)f'(x).
  3. f(x)=5tan⁡xf(x) = 5\tan x. Find f′(x)f'(x).

Watch out!

Practice ★★

  1. f(x)=sec⁡x+2cot⁡xf(x) = \sec x + 2\cot x. Find f′(x)f'(x).
  2. f(x)=sec⁡x+8cot⁡xf(x) = \sec x + 8\cot x. Find f′(x)f'(x).
  1. Find the slope of the tangent line to y=tan⁡xy = \tan x at x=π6x = \dfrac{\pi}{ 6 }. Give an exact value.
  2. Find the slope of the tangent line to y=tan⁡xy = \tan x at x=π4x = \dfrac{\pi}{ 4 }. Give an exact value.

Exit ticket

Three questions to check you have got it.

  1. f(x)=7tan⁡xf(x) = 7\tan x. Find f′(x)f'(x).
  2. f(x)=3tan⁡xf(x) = 3\tan x. Find f′(x)f'(x).
  3. f(x)=5tan⁡xf(x) = 5\tan x. Find f′(x)f'(x).

Summary card

tan, sec, cot, csc

  • (tan⁡x)′=sec⁡2x(\tan x)' = \sec^2 x, (sec⁡x)′=sec⁡xtan⁡x(\sec x)' = \sec x\tan x.
  • (cot⁡x)′=−csc⁡2x(\cot x)' = -\csc^2 x, (csc⁡x)′=−csc⁡xcot⁡x(\csc x)' = -\csc x\cot x.

Watch out: co-functions take a minus sign. You may also write sec⁡2x\sec^2 x as 1cos⁡2x\dfrac{1}{\cos^2 x}.

7. The chain rule

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is y=sin⁡xy = \sin x; the blue one is y=sin⁡(kx)y = \sin(kx). Increase kk.

The curves y = sin x and y = sin(kx). With k = 2 the blue curve oscillates twice as fast, and its steepest slope is 2 instead of 1.

At k=2k = 2 the blue curve goes twice as fast. At x=0x = 0, what is the slope of y=sin⁡(2x)y = \sin(2x)?

  1. 2
  2. 1

Key idea

For a function inside a function: differentiate the outside, keep the inside, multiply by the derivative of the inside.

Chain rule: (f(g(x)))′=f′(g(x))⋅g′(x)\big(f(g(x))\big)' = f'(g(x)) \cdot g'(x).

Worked example

h(x)=(x2+1)5h(x) = (x^2 + 1)^5. Find h′(x)h'(x).

f(u)=u5f(u) = u^5, g(x)=x2+1g(x) = x^2 + 1
Outside u⁵, inside u = x² + 1.
h′(x)=5(x2+1)4⋅2xh'(x) = 5(x^2 + 1)^4 \cdot 2x
f'(u) = 5u⁴ with the inside kept; g'(x) = 2x.
h′(x)=10x(x2+1)4h'(x) = 10x(x^2 + 1)^4
Tidy.

Your turn, guided

Your turn: h(x)=e3xh(x) = e^{3x}.

h′(x)=3e3xh'(x) = 3e^{3x}
Outside eᵘ, inside 3x: eᵘ times 3.

Practice ★

  1. f(x)=(6x+7)4f(x) = (6x + 7)^{ 4 }. Find f′(x)f'(x).
  2. f(x)=(7x+6)4f(x) = (7x + 6)^{ 4 }. Find f′(x)f'(x).
  3. f(x)=(3x+5)4f(x) = (3x + 5)^{ 4 }. Find f′(x)f'(x).

Watch out!

Practice ★★

  1. f(x)=sin⁡(4x2)f(x) = \sin(4x^2). Find f′(x)f'(x).
  2. f(x)=sin⁡(6x2)f(x) = \sin(6x^2). Find f′(x)f'(x).
  1. f(x)=e5x2f(x) = e^{ 5x^2}. Find f′(x)f'(x).
  2. f(x)=e4x2f(x) = e^{ 4x^2}. Find f′(x)f'(x).
  1. Boss problem. h(x)=x3sin⁡(4x)h(x) = x^{ 3 }\sin(4x). Find the slope of its graph at x=π4x = \dfrac{\pi}{ 4 }.

Exit ticket

Three questions to check you have got it.

  1. f(x)=(6x+7)4f(x) = (6x + 7)^{ 4 }. Find f′(x)f'(x).
  2. f(x)=(7x+6)4f(x) = (7x + 6)^{ 4 }. Find f′(x)f'(x).
  3. f(x)=(3x+5)4f(x) = (3x + 5)^{ 4 }. Find f′(x)f'(x).

Summary card

Chain rule

(f(g(x)))′=f′(g(x)) g′(x)\big(f(g(x))\big)' = f'(g(x))\,g'(x): outside's derivative (inside kept) times inside's derivative.

Watch out: do not forget the factor g′(x)g'(x).

8. Derivatives of inverse functions: ln x, arcsin x, arctan x

Warm-up

Three quick questions from earlier lessons.

Discover

y=ln⁡xy = \ln x means ey=xe^y = x. The slope of eye^y is ey=xe^y = x. The graphs of ln⁡\ln and exp⁡\exp are reflections, so their slopes are reciprocal. What is (ln⁡x)′(\ln x)'?

  1. 1x\dfrac{1}{x}
  2. ln⁡x\ln x

Key idea

The derivatives of the inverse functions come from the chain rule.

(ln⁡x)′=1x(\ln x)' = \dfrac{1}{x} (x>0x > 0); (arcsin⁡x)′=11−x2(\arcsin x)' = \dfrac{1}{\sqrt{1 - x^2}}; (arctan⁡x)′=11+x2(\arctan x)' = \dfrac{1}{1 + x^2}. With an inside function: (ln⁡u)′=u′u(\ln u)' = \dfrac{u'}{u}.

Worked example

h(x)=ln⁡(x3+2)h(x) = \ln(x^3 + 2). Find h′(x)h'(x).

(ln⁡u)′=u′u(\ln u)' = \dfrac{u'}{u}
Inside u = x³ + 2, u' = 3x².
h′(x)=3x2x3+2h'(x) = \dfrac{3x^2}{x^3 + 2}
Substitute.

Your turn, guided

Your turn: h(x)=arctan⁡(x2)h(x) = \arctan(x^2).

h′(x)=2x1+x4h'(x) = \dfrac{2x}{1 + x^4}
u = x², u' = 2x; divide by 1 + u² = 1 + x⁴.

Practice ★

  1. f(x)=ln⁡(9x)f(x) = \ln(9x) for x>0x > 0. Find f′(x)f'(x).
  2. f(x)=ln⁡(2x)f(x) = \ln(2x) for x>0x > 0. Find f′(x)f'(x).
  1. f(x)=ln⁡(x2+2)f(x) = \ln(x^2 + 2). Find f′(x)f'(x).
  2. f(x)=ln⁡(x2+8)f(x) = \ln(x^2 + 8). Find f′(x)f'(x).

Watch out!

Practice ★★

  1. f(x)=arctan⁡(7x)f(x) = \arctan(7x). Find f′(x)f'(x).
  2. f(x)=arctan⁡(2x)f(x) = \arctan(2x). Find f′(x)f'(x).

Exit ticket

Three questions to check you have got it.

  1. f(x)=ln⁡(x2+2)f(x) = \ln(x^2 + 2). Find f′(x)f'(x).
  2. f(x)=ln⁡(x2+8)f(x) = \ln(x^2 + 8). Find f′(x)f'(x).
  3. f(x)=ln⁡(x2+1)f(x) = \ln(x^2 + 1). Find f′(x)f'(x).

Summary card

Inverse functions

  • (ln⁡x)′=1x(\ln x)' = \dfrac{1}{x}, (ln⁡u)′=u′u(\ln u)' = \dfrac{u'}{u}.
  • (arcsin⁡x)′=11−x2(\arcsin x)' = \dfrac{1}{\sqrt{1 - x^2}}, (arctan⁡x)′=11+x2(\arctan x)' = \dfrac{1}{1 + x^2}.

Watch out: the chain rule still applies to the inside.