The definition of the derivative, product and quotient rules

The derivative means “instantaneous rate of change” and “slope of the tangent line”. This week you learn two things: computing the derivative from the definition (as a limit) and computing it quickly with rules. On exams they are asked separately. If the question says “Use the definition”, an answer found with a rule gets no points. The key points:

1. Definition: $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$. It always gives $\tfrac00$, and you cancel $h$ with algebra.
2. Tangent line: $y-f(a)=f'(a)(x-a)$.
3. Write roots and fractions as powers first, then use the power rule.
4. The derivative of a product is not the product of the derivatives. $(fg)'=f'g+fg'$.
5. In the quotient rule, order matters: $\dfrac{f'g-fg'}{g^2}$.

The definition of the derivative

The derivative of $f$ at the point $a$:

$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$

The fraction is the slope of the secant line through the points $a$ and $a+h$. As $h$ goes to zero, the secant line turns into the tangent line. If the limit exists, $f$ is differentiable at $a$.

Notations: $f'(x)$, $y'$, $\dfrac{dy}{dx}$, $\dfrac{d}{dx}f(x)$.

Computing from the definition: three types

Example 1 (polynomial). $f(x)=x^2$, $a=3$.

f(3+h) = 9 + 6h + h^2
f(3+h) - f(3) = 6h + h^2
[ ... ] / h = 6 + h
h -> 0:  f'(3) = 6

The tangent line passes through $(3,9)$ with slope $6$: $y-9=6(x-3)$, that is $y=6x-9$.

Example 2 (fraction). $f(x)=\dfrac1x$. Find a common denominator:

$\frac{1}{h}\left(\frac{1}{x+h}-\frac1x\right)=\frac1h\cdot\frac{x-(x+h)}{x(x+h)}=\frac{-1}{x(x+h)}\ \xrightarrow{h\to0}\ -\frac{1}{x^2}$

Example 3 (root). $f(x)=\sqrt x$. Multiply by the conjugate:

$\frac{\sqrt{x+h}-\sqrt x}{h}\cdot\frac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}=\frac{h}{h(\sqrt{x+h}+\sqrt x)}\ \xrightarrow{h\to0}\ \frac{1}{2\sqrt x}$

PITFALL · setting $h$ to zero too early. If you put $h=0$ before $h$ has completely left the denominator, you get $\tfrac00$. Simplify first, then take the limit.

The derivative is also a function

When you put the variable $x$ in place of $a$, $f'(x)$ becomes a new function: it assigns to each point the slope at that point. That is why you can read the graph of the derivative from the graph of $f$:

• if $f$ is increasing, $f'>0$ (the graph is above the axis),
• if $f$ is decreasing, $f'<0$,
• at a peak or valley of $f$ there is a horizontal tangent, $f'=0$.

The exam question “which of the following graphs is the graph of $f'$?” is solved with these three sentences. In week 9 you will use the same idea in reverse and sketch the shape of $f$ from the sign of $f'$.

There is also a difference between average and instantaneous change. The average rate of change on the interval $[a,b]$ is $\dfrac{f(b)-f(a)}{b-a}$, the slope of the secant. The instantaneous rate of change is $f'(a)$. In physics, this is the difference between average velocity and the velocity at that moment.

Differentiability and continuity

Theorem: If $f$ is differentiable at $a$, it is continuous at $a$. The converse is not true.

The function $f(x)=|x|$ is continuous at $0$. But

$\lim_{h\to0^+}\frac{|h|}{h}=1,\qquad \lim_{h\to0^-}\frac{|h|}{h}=-1$

The one-sided derivatives differ, so $f'(0)$ does not exist. The graph has a corner.

Typical places where the derivative does not exist: a corner, a cusp (for example $x^{2/3}$ at $0$), a vertical tangent ($\sqrt[3]{x}$ at $0$), a point of discontinuity.

Differentiation rules

Example 4 (power rule). $y=3x^4-\dfrac2x+5\sqrt[3]{x^2}$.

First write it as powers: $y=3x^4-2x^{-1}+5x^{2/3}$. Then:

$y'=12x^3+2x^{-2}+\frac{10}{3}x^{-1/3}=12x^3+\frac{2}{x^2}+\frac{10}{3\sqrt[3]{x}}$

Example 5 (product). $y=(2x+1)(x^2-3)$.

$y'=\underbrace{2}_{f'}(x^2-3)+(2x+1)\underbrace{2x}_{g'}=2x^2-6+4x^2+2x=6x^2+2x-6$

Example 6 (quotient). $y=\dfrac{x^2}{x+1}$.

$y'=\frac{2x(x+1)-x^2\cdot1}{(x+1)^2}=\frac{x^2+2x}{(x+1)^2}$

To remember the quotient rule: “bottom times derivative of top, minus top times derivative of bottom, over bottom squared”.

Where do the rules come from?

The power rule comes from applying the binomial expansion to the definition: $(x+h)^n=x^n+nx^{n-1}h+(\ldots)$, where the part in parentheses contains terms with $h^2$ and higher powers. When the difference is divided by $h$, $nx^{n-1}$ remains, and every other term goes to zero because it contains $h$.

For the product rule, in the definition you add and subtract the term $f(x+h)g(x)$:

$\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=f(x+h)\frac{g(x+h)-g(x)}{h}+g(x)\frac{f(x+h)-f(x)}{h}$

As $h\to0$ the first term becomes $f(x)g'(x)$ and the second becomes $g(x)f'(x)$. (The part $f(x+h)\to f(x)$ comes from a differentiable function being continuous.) The proof is rarely asked on exams, but this is the answer to “why not $f'g'$”.

For three factors the rule extends with the same logic: each time differentiate one factor and leave the others as they are.

$(fgh)'=f'gh+fg'h+fgh'$

Example 7. For $y=x(x+1)(x-2)$, $y'=(x+1)(x-2)+x(x-2)+x(x+1)=3x^2-2x-2$. To check, expand: $y=x^3-x^2-2x$, $y'=3x^2-2x-2$. The same.

PITFALL · an unnecessary quotient rule. For $y=\dfrac{x^3+1}{x}$, instead of the quotient rule, divide first: $y=x^2+x^{-1}$, $y'=2x-x^{-2}$. It is shorter and has less risk of error. If the denominator is a constant ($\tfrac{x^2}{5}$), never use the quotient rule.

Higher-order derivatives

$y''=\dfrac{d^2y}{dx^2}$ is the derivative of the derivative. $y'''$, $y^{(4)}$, … continue in this way.

Example 8. $y=x^4-2x^3$: $y'=4x^3-6x^2$, $y''=12x^2-12x$, $y'''=24x-12$, $y^{(4)}=24$, $y^{(5)}=0$. For a polynomial each derivative lowers the degree by one; the $(n+1)$th derivative of a polynomial of degree $n$ is zero.

Physical meaning: velocity and acceleration

If the position is $s(t)$, the velocity is $v(t)=s'(t)$ and the acceleration is $a(t)=v'(t)=s''(t)$. The sign of the velocity gives the direction: $v>0$ forward, $v<0$ backward. Speed is $|v|$.

Example 9. $s(t)=t^3-6t^2+9t$ (meters, seconds). $v(t)=3t^2-12t+9$, $a(t)=6t-12$. At $t=2$, $v=12-24+9=-3$ m/s (moving backward), $a=0$. The object stops when $v=0$, that is $3(t-1)(t-3)=0 \Rightarrow t=1$ and $t=3$.

Tangent and normal lines

At the point $x=a$:

• Tangent: $y-f(a)=f'(a)(x-a)$
• Normal (perpendicular to the tangent): $y-f(a)=-\dfrac{1}{f'(a)}(x-a)$, if $f'(a)\ne0$

The question “where is the tangent horizontal?” is the equation $f'(x)=0$. For “tangent parallel to a given line”, you equate the slopes.

Example 10. At which point of the curve $y=x^2-3x+4$ is the tangent parallel to the line $y=5x-2$?

parallel -> equal slopes:  y' = 2x - 3 = 5   ->  x = 4
point:  y(4) = 16 - 12 + 4 = 8    ->  (4, 8)
tangent:  y - 8 = 5(x - 4)        ->  y = 5x - 12

Example 11. The normal line to the curve $y=\sqrt x$ at $x=4$. The point is $(4,2)$. The tangent slope is $\tfrac{1}{2\sqrt4}=\tfrac14$ and the normal slope is $-4$. Normal: $y-2=-4(x-4)$, that is $y=-4x+18$.

Common mistakes