The Derivative

Study Sheet

The Derivative

Definition, differentiability, and the basic rules of differentiation

Rates of Change and the Tangent Line

Concept
Average vs. instantaneous rate of change

The average rate of change of ff on [a,b][a,b] is the slope of the secant line through the endpoints:

f(b)−f(a)b−a.\frac{f(b)-f(a)}{b-a}.

The instantaneous rate of change at x=ax=a is the slope of the tangent line at that point. It is the limit of the average rates as the interval shrinks to a single point, and we call it the derivative f′(a)f'(a).

The tangent line at x=ax=a touches the curve at (a,f(a))\big(a,f(a)\big) and has slope f′(a)f'(a). Its equation is

y=f(a)+f′(a) (x−a).y = f(a) + f'(a)\,(x-a).
Example
Worked example: average rate of change

Find the average rate of change of f(x)=x2f(x)=x^2 on [1,3][1,3].

f(3)−f(1)3−1=9−12=82=4.\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=\frac{8}{2}=4.

So a secant line over this interval has slope 44.

Tip

Tip: “Average” ⇒\Rightarrow secant ⇒\Rightarrow two points. “Instantaneous” ⇒\Rightarrow tangent ⇒\Rightarrow one point (a limit).

The Limit Definition of the Derivative

Concept
Definition

The derivative of ff as a function is

f′(x)=lim⁡h→0f(x+h)−f(x)h,f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h},

provided the limit exists. The derivative at a point x=ax=a is

f′(a)=lim⁡h→0f(a+h)−f(a)h.f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}.

The alternate form of the derivative at aa is

f′(a)=lim⁡x→af(x)−f(a)x−a.f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.

Notations: f′(x)f'(x), dydx\dfrac{dy}{dx}, ddx[f(x)]\dfrac{d}{dx}\big[f(x)\big], and y′y' all mean the derivative.

Example
Worked example: derivative from the definition

Let f(x)=x2f(x)=x^2. Then

f′(x)=lim⁡h→0(x+h)2−x2h=lim⁡h→0x2+2xh+h2−x2h=lim⁡h→02xh+h2h=lim⁡h→0(2x+h)=2x.f'(x)=\lim_{h\to0}\frac{(x+h)^2-x^2}{h} =\lim_{h\to0}\frac{x^2+2xh+h^2-x^2}{h} =\lim_{h\to0}\frac{2xh+h^2}{h}=\lim_{h\to0}(2x+h)=2x.
Tip

Tip: The whole point of algebra in these problems is to cancel the hh in the denominator before letting h→0h\to0; otherwise you get 00\tfrac{0}{0}.

Differentiability and Continuity

Concept
When does the derivative fail to exist?

If ff is differentiable at aa, then ff is continuous at aa. The converse is false: a continuous function need not be differentiable. A function fails to be differentiable at a point where its graph has a

  • [leftmargin=1.4em,itemsep=1pt,topsep=1pt]
  • corner (slopes from the two sides disagree), e.g. y=∣x∣y=|x| at x=0x=0;
  • cusp (slopes head toward +∞+\infty and −∞-\infty), e.g. y=x2/3y=x^{2/3} at x=0x=0;
  • vertical tangent (slope is infinite), e.g. y=x1/3y=x^{1/3} at x=0x=0;
  • discontinuity (jump, hole, or asymptote).

 

 

Example
Worked example: a corner

f(x)=∣x∣f(x)=|x| is continuous at x=0x=0, but the slope from the left is −1-1 and the slope from the right is +1+1. Since these one-sided derivatives disagree, f′(0)f'(0) does not exist.

Tip

Tip: “Differentiable ⇒\Rightarrow continuous” is a one-way street. To disprove differentiability, either show a break (not continuous) or show the left- and right-hand slopes differ.

Power, Constant, Constant-Multiple, and Sum/Difference Rules

Example
Worked example: power and sum rules
ddx[3x4−2x2+7]=3(4x3)−2(2x)+0=12x3−4x.\frac{d}{dx}\big[3x^4-2x^2+7\big]=3(4x^3)-2(2x)+0=12x^3-4x.

Rewrite roots and fractions as powers first:

ddx[x]=ddx[x1/2]=12x−1/2=12x.\frac{d}{dx}\Big[\sqrt{x}\Big]=\frac{d}{dx}\big[x^{1/2}\big]=\tfrac12 x^{-1/2}=\frac{1}{2\sqrt{x}}.
Tip

Tip: Before differentiating, rewrite 1xk\dfrac{1}{x^k} as x−kx^{-k} and xkm\sqrt[m]{x^k} as xk/mx^{k/m} so the power rule applies directly.

Reminder — The differentiation rules:  (xn)′=nxn−1,(fg)′=f′g+fg′,(fg)′=f′g−fg′g2,(f(g(x)))′=f′(g(x))g′(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)

Derivatives of Exponential, Logarithmic, and Trig Functions

Example
Worked example: combine with the basic rules
ddx[3sin⁡x−2cos⁡x+ex]=3cos⁡x−2(−sin⁡x)+ex=3cos⁡x+2sin⁡x+ex.\frac{d}{dx}\big[3\sin x-2\cos x+e^x\big]=3\cos x-2(-\sin x)+e^x=3\cos x+2\sin x+e^x.
Tip

Tip: The two “co-” functions (cos⁡\cos, cot⁡\cot, csc⁡\csc) each pick up a minus sign when differentiated. Watch signs carefully.

Product Rule, Quotient Rule, and Higher-Order Derivatives

Concept
Products and quotients
(fg)′=f′g+fg′⏟Product Rule(fg)′=f′g−fg′g2⏟Quotient Rule\underbrace{(fg)'=f'g+fg'}_{\text{Product Rule}} \qquad\qquad \underbrace{\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}}_{\text{Quotient Rule}}

Higher-order derivatives come from differentiating again: f′′=(f′)′f''=(f')', f′′′=(f′′)′f'''=(f'')', and so on. Notation: f′′(x)f''(x), d2ydx2\dfrac{d^2y}{dx^2}.

Example
Worked example: product and quotient rules

Product:

ddx[x2sin⁡x]=(2x)sin⁡x+x2cos⁡x=2xsin⁡x+x2cos⁡x.\frac{d}{dx}\big[x^2\sin x\big]=(2x)\sin x+x^2\cos x=2x\sin x+x^2\cos x.

Quotient:

ddx ⁣[sin⁡xx]=(cos⁡x)(x)−(sin⁡x)(1)x2=xcos⁡x−sin⁡xx2.\frac{d}{dx}\!\left[\frac{\sin x}{x}\right]=\frac{(\cos x)(x)-(\sin x)(1)}{x^2}=\frac{x\cos x-\sin x}{x^2}.

Interpreting f′f' and f′′f''

The sign of f′f' tells you whether ff is increasing (f′>0f'>0) or decreasing (f′<0f'<0); where f′=0f'=0 and changes sign, ff has a local max or min. The sign of f′′f'' tells you concavity (f′′>0f''>0 concave up, f′′<0f''<0 concave down). Below, f(x)=x2−1f(x)=x^2-1 (blue) has a minimum exactly where its derivative f′(x)=2xf'(x)=2x (red) crosses zero.

Tip

Tip: For a quotient, some students remember “low dd-high minus high dd-low, over low squared.” Never divide the two derivatives−-always use the full formula.

Going Deeper: Advanced Derivative Ideas

Concept
1. Derivative of x\sqrt{x} from the definition

Multiply by the conjugate to clear the root:

f′(x)=lim⁡h→0x+h−xh⋅x+h+xx+h+x=lim⁡h→0(x+h)−xh(x+h+x)=lim⁡h→01x+h+x=12x.f'(x)=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} =\lim_{h\to0}\frac{(x+h)-x}{h\big(\sqrt{x+h}+\sqrt{x}\big)} =\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}.

This agrees with the power rule on x1/2x^{1/2}.

Concept
2. Derivative of 1x\dfrac{1}{x} from the definition

Combine the fractions in the numerator first:

f′(x)=lim⁡h→01x+h−1xh=lim⁡h→0x−(x+h)x(x+h)h=lim⁡h→0−hh x(x+h)=lim⁡h→0−1x(x+h)=−1x2.f'(x)=\lim_{h\to0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h} =\lim_{h\to0}\frac{\frac{x-(x+h)}{x(x+h)}}{h} =\lim_{h\to0}\frac{-h}{h\,x(x+h)} =\lim_{h\to0}\frac{-1}{x(x+h)}=-\frac{1}{x^2}.

Again this matches the power rule on x−1x^{-1}.

Example
3. A harder definition problem: f(x)=1xf(x)=\dfrac{1}{\sqrt{x}}
f′(x)=lim⁡h→01x+h−1xh=lim⁡h→0x−x+hhxx+h⋅x+x+hx+x+h=lim⁡h→0−1xx+h(x+x+h)=−12x3/2.f'(x)=\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} =\lim_{h\to0}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}\cdot\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} =\lim_{h\to0}\frac{-1}{\sqrt{x}\sqrt{x+h}\big(\sqrt{x}+\sqrt{x+h}\big)}=-\frac{1}{2x^{3/2}}.

Consistent with the power rule on x−1/2x^{-1/2}: −12x−3/2-\tfrac12 x^{-3/2}.

Concept
4. One-sided derivatives and a piecewise function

The one-sided derivatives at aa are

f−′(a)=lim⁡h→0−f(a+h)−f(a)h,f+′(a)=lim⁡h→0+f(a+h)−f(a)h.f'_-(a)=\lim_{h\to0^-}\frac{f(a+h)-f(a)}{h},\qquad f'_+(a)=\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}.

ff is differentiable at aa exactly when both exist and are equal. This is how you solve for parameters.

Example
4b. Solving for parameters

Find a,ba,b so that

f(x)={x2,x≤1,ax+b,x>1,is differentiable at x=1.f(x)=\begin{cases} x^2, & x\le 1,\\[2pt] ax+b, & x>1,\end{cases} \qquad\text{is differentiable at } x=1.

Continuity at x=1x=1: the pieces must match, 12=a(1)+b1^2=a(1)+b, so a+b=1a+b=1. Matching slopes at x=1x=1: derivative of x2x^2 is 2x2x, giving 22 at x=1x=1; derivative of ax+bax+b is aa. So a=2a=2, and then b=1−a=−1b=1-a=-1. Thus a=2, b=−1a=2,\ b=-1.

Concept
5. Proving the Product Rule

Add and subtract f(x+h)g(x)f(x+h)g(x) in the numerator:

(fg)′(x)=lim⁡h→0f(x+h)g(x+h)−f(x)g(x)h=lim⁡h→0f(x+h)[g(x+h)−g(x)]+g(x)[f(x+h)−f(x)]h=lim⁡h→0f(x+h)⋅g(x+h)−g(x)h+g(x)⋅lim⁡h→0f(x+h)−f(x)h=f(x)g′(x)+g(x)f′(x).\begin{aligned} (fg)'(x)&=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=\lim_{h\to0}\frac{f(x+h)\big[g(x+h)-g(x)\big]+g(x)\big[f(x+h)-f(x)\big]}{h}\\ &=\lim_{h\to0}f(x+h)\cdot\frac{g(x+h)-g(x)}{h}+g(x)\cdot\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &=f(x)g'(x)+g(x)f'(x). \end{aligned}

(We used that ff is continuous, so f(x+h)→f(x)f(x+h)\to f(x).)

Concept
6. Proving the Quotient Rule

Let Q=fgQ=\dfrac{f}{g}, so f=Qgf=Qg. Differentiate both sides with the product rule: f′=Q′g+Qg′f'=Q'g+Qg'. Solve for Q′Q' and substitute Q=fgQ=\dfrac{f}{g}:

Q′=f′−Qg′g=f′−fgg′g=f′g−fg′g2.Q'=\frac{f'-Qg'}{g}=\frac{f'-\frac{f}{g}g'}{g}=\frac{f'g-fg'}{g^2}.
Concept
7. The symmetric difference quotient

Calculators often estimate f′(x)f'(x) with the symmetric difference quotient

f′(x)≈f(x+h)−f(x−h)2h.f'(x)\approx\frac{f(x+h)-f(x-h)}{2h}.

It uses points on both sides and is usually more accurate. For f(x)=x2f(x)=x^2 it is even exact for every hh:

(x+h)2−(x−h)22h=4xh2h=2x.\frac{(x+h)^2-(x-h)^2}{2h}=\frac{4xh}{2h}=2x.

Warning: this quotient can return a finite value even where f′f' does not exist. For f(x)=∣x∣f(x)=|x| at x=0x=0 it gives ∣h∣−∣−h∣2h=0\frac{|h|-|-h|}{2h}=0, yet f′(0)f'(0) truly does not exist−-so a numerical value alone never proves differentiability.

Formulas, Proofs & Tips

Tip
The difference quotient and the derivative
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

What it means. The slope of the tangent line — the instantaneous rate of change.

Example. For f(x)=x2f(x)=x^2: f′(x)=lim⁡h→0(x+h)2−x2h=2xf'(x)=\lim_{h\to0}\tfrac{(x+h)^2-x^2}{h}=2x.

Why it works. f(x+h)−f(x)h\tfrac{f(x+h)-f(x)}{h} is the slope of the secant through two nearby points. Letting h→0h\to 0 slides the second point into the first, so the secant becomes the tangent.

Tip. The hh must cancel before you substitute h=0h=0; otherwise you get the meaningless 00\tfrac00.

Tip
The differentiation rules
(xn)′=nxn−1,(fg)′=f′g+fg′,(fg)′=f′g−fg′g2,(f(g(x)))′=f′(g(x))g′(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)

What it means. Power, product, quotient and chain rules — enough to differentiate almost anything.

Example. (x3)′=3x2(x^3)'=3x^2, and (x2sin⁡x)′=2xsin⁡x+x2cos⁡x(x^2\sin x)'=2x\sin x+x^2\cos x.

Why it works. The product rule comes from the difference quotient of fgfg: add and subtract f(x+h)g(x)f(x+h)g(x) in the numerator to split it into a piece that becomes f′gf'g and a piece that becomes fg′fg'. The chain rule multiplies rates: if uu changes g′g' times as fast as xx, and ff changes f′f' times as fast as uu, the combined rate is the product.

Tip. Product rule is not f′g′f'g'. For the chain rule, always finish with the derivative of the inside.

Tip
The Mean Value Theorem
f′(c)=f(b)−f(a)b−afor some c∈(a,b)f'(c)=\frac{f(b)-f(a)}{b-a} \quad\text{for some } c \in (a,b)

What it means. Somewhere in the interval the instantaneous rate equals the average rate.

Example. For f(x)=x2f(x)=x^2 on [0,2][0,2]: f′(c)=4−02−0=2f'(c)=\tfrac{4-0}{2-0}=2, which occurs at c=1c=1.

Why it works. Tilt the graph so the endpoints are level (subtract the secant line). The resulting function has equal endpoint values, so it has an interior max or min, where the derivative is 00 — untilting gives f′(c)f'(c) equal to the secant slope.

Tip. Needs ff continuous on [a,b][a,b] and differentiable on (a,b)(a,b). Rolle's Theorem is the special case f(a)=f(b)f(a)=f(b).