The Algebra of Derivatives

Once we know the derivatives of basic functions, we need rules for differentiating their sums, differences, products, and quotients. This section establishes four fundamental rules and shows how to apply them together.

Rule	Formula
Constant Multiple	(cu)'=cu'
Sum	(u+v)'=u'+v'
Product	(uv)'=u'v+uv'
Quotient	\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}

Constant Multiple Rule and Sum Rule

Assume $f$ and $g$ are differentiable at $x$ .

The Constant Multiple Rule. If $ϕ (x) = c f (x)$ where $c$ is a constant, then \phi'(x)=(cf)'(x)=cf'(x)
The Sum Rule. If $ϕ (x) = f (x) + g (x)$ , then \phi'(x)=(f+g)'(x)=f'(x)+g'(x)

These two rules say:

The derivative of a constant times a function equals the constant times the derivative of the function.
The derivative of the sum of two functions equals the sum of their individual derivatives.

If $u = f (x)$ and $v = g (x)$ , the rules can be written as

(cu)'=cu'\qquad\text{or}\qquad\frac{d}{dx}(cu)=c\frac{du}{dx}(u+v)'=u'+v'\qquad\text{or}\qquad\frac{d}{dx}(u+v)=\frac{du}{dx}+\frac{dv}{dx}

Proof

**Constant Multiple Rule.** \phi'(x)=\lim_{\Delta x\to0}\frac{cf(x+\Delta x)-cf(x)}{\Delta x}=c\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=cf'(x) **Sum Rule.** \begin{aligned} \phi'(x)&=\lim_{\Delta x\to0}\frac{(f+g)(x+\Delta x)-(f+g)(x)}{\Delta x}\\ &=\lim_{\Delta x\to0}\left[\frac{f(x+\Delta x)-f(x)}{\Delta x}+\frac{g(x+\Delta x)-g(x)}{\Delta x}\right]\\ &=f'(x)+g'(x) \end{aligned}

Find $\frac{d z}{d t}$ if $z = 4 \sqrt{t}$ .

Solution

Since

z = 4 t^{1 / 2}

\frac{d z}{d t} = 4 \cdot \frac{1}{2} t^{1 / 2 - 1} = 2 t^{- 1 / 2} = \frac{2}{\sqrt{t}}

Differentiate $f (x) = 5 x^{4} - \sqrt{x}$ .

Solution

\begin{aligned} f'(x)&=(5x^4)'-(\sqrt{x})'\\ &=5(x^4)'-\left(x^{1/2}\right)'\\ &=5\cdot4x^3-\frac{1}{2}x^{-1/2}\\ &=20x^3-\frac{1}{2\sqrt{x}} \end{aligned}

Find $\frac{d y}{d u}$ if $y = 0.2 u^{0.8} - \frac{3}{4 u} + \sqrt[3]{u} + 9$ .

Solution

\begin{aligned} \frac{dy}{du}&=0.2\cdot0.8\,u^{-0.2}-\frac{3}{4}\cdot(-1)u^{-2}+\frac{1}{3}u^{-2/3}+0\\ &=\frac{0.16}{u^{0.2}}+\frac{3}{4u^2}+\frac{1}{3\sqrt[3]{u^2}} \end{aligned}

Additional results from the Sum and Constant Multiple Rules:

Difference Rule: (f-g)'(x)=f'(x)-g'(x).
Derivative of a constant: $\frac{d c}{d x} = 0$ .
Linearity: $\frac{d}{d x} \sum_{i = 1}^{n} c_{i} f_{i} (x) = \sum_{i = 1}^{n} c_{i} \frac{d f_{i} (x)}{d x}$ .

Product Rule

The Product Rule. Assume $f$ and $g$ are differentiable at $x$ . If $ϕ (x) = f (x) g (x)$ , then

\phi'(x)=(fg)'(x)=f'(x)g(x)+f(x)g'(x)

In terms of $u = f (x)$ and $v = g (x)$ :

(uv)'=u'v+v'u\qquad\text{or}\qquad\frac{d}{dx}(uv)=\frac{du}{dx}v+u\frac{dv}{dx}

Proof

\phi'(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x} Adding and subtracting

f (x) g (x + Δ x)

in the numerator: \begin{aligned} \phi'(x)&=\lim_{\Delta x\to0}\left[g(x+\Delta x)\frac{f(x+\Delta x)-f(x)}{\Delta x}+f(x)\frac{g(x+\Delta x)-g(x)}{\Delta x}\right] \end{aligned} Since

g

is continuous (differentiable implies continuous),

g (x + Δ x) \to g (x)

Δ x \to 0

, giving \phi'(x)=g(x)f'(x)+f(x)g'(x)

Quotient Rule

The Quotient Rule. If $ϕ (x) = \frac{f (x)}{g (x)}$ and $g (x) \neq 0$ , then

\phi'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}

In differential notation:

\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\qquad\text{or}\qquad\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{\dfrac{du}{dx}v-u\dfrac{dv}{dx}}{v^2}

Proof

Let

y = u / v

. When

x

changes by

Δ x

Δ y = \frac{u + Δ u}{v + Δ v} - \frac{u}{v} = \frac{v Δ u - u Δ v}{v (v + Δ v)}

Dividing by

Δ x

and taking

Δ x \to 0

: \frac{dy}{dx}=\frac{v\,\dfrac{du}{dx}-u\,\dfrac{dv}{dx}}{v^2}=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}

Important note: The order in the numerator of the Quotient Rule matters because of the minus sign. Unlike the Product Rule, f'g-fg' is not the same as fg'-f'g.

Worked Examples

Given $y = x^{2}$ , differentiate using the Product Rule.

Solution

Write

y = f (x) g (x)

with

f (x) = x

and

g (x) = x

: y'=f'(x)g(x)+f(x)g'(x)=1\cdot x+x\cdot1=2x

Find the derivative of $y = 4 x^{3} - 2 x^{2} + 6 x + 5$ .

Solution

\begin{aligned} \frac{dy}{dx}&=\frac{d}{dx}(4x^3)-\frac{d}{dx}(2x^2)+\frac{d}{dx}(6x)+\frac{d}{dx}5\\ &=4(3x^2)-2(2x)+6\cdot1+0\\ &=12x^2-4x+6 \end{aligned}

Prove $\frac{d}{d x} x^{n} = n x^{n - 1}$ where $n$ is a negative integer.

Solution

Let

n = - m

where

m > 0

. Then

x^{n} = x^{- m} = \frac{1}{x^{m}}

. Using the Quotient Rule: \begin{aligned} \frac{d}{dx}\frac{1}{x^m}&=\frac{0\cdot x^m-mx^{m-1}\cdot1}{(x^m)^2}=-m\frac{x^{m-1}}{x^{2m}}=-mx^{m-1-2m}=nx^{n-1} \end{aligned}

Given $y = 1 / x^{3}$ , find $d y / d x$ .

Solution

y = x^{- 3}

, so

\frac{d}{d x} y = - 3 x^{- 4} = - \frac{3}{x^{4}}

Given $y = \sqrt{x}$ ( $x > 0$ ), find $d y / d x$ .

Solution

Let

f (x) = \sqrt{x}

and

g (x) = f (x) f (x) = x

. By the Product Rule g'(x)=2f(x)f'(x)=1, so f'(x)=\dfrac{1}{2\sqrt{x}}.

Find $d y / d x$ if $y = x^{1 / n}$ where $n$ is an integer.

Solution

Let

g (x) = [f (x)]^{n} = x

where

f (x) = x^{1 / n}

. Using the Product Rule

n

times: g'(x)=n[f(x)]^{n-1}f'(x)=1 Therefore f'(x)=\dfrac{1}{nx^{(n-1)/n}}=\dfrac{1}{n}x^{1/n-1}.

Differentiate $f (x) = \frac{x^{2} - 1}{x^{3} + 1}$ .

Solution

\begin{aligned} f'(x)&=\frac{2x(x^3+1)-(x^2-1)(3x^2)}{(x^3+1)^2}\\ &=\frac{2x^4+2x-3x^4+3x^2}{(x^3+1)^2}\\ &=\frac{-x^4+3x^2+2x}{(x^3+1)^2} \end{aligned}

Prove $\frac{d}{d x} u^{n} = n u^{n - 1} \frac{d u}{d x}$ where $n$ is a positive integer.

Solution

Let

y = u^{n}

. Then

\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x} = n u^{n - 1} \frac{d u}{d x}

Frequently Asked Questions

How do I remember the Quotient Rule?

A common mnemonic is "low d-high minus high d-low, over low squared." That is: denominator times derivative of numerator, minus numerator times derivative of denominator, all divided by denominator squared.

Can the Product Rule be extended to three or more functions?

Yes. If

y = f_{1} f_{2} \dots f_{n}

then y' is the sum of

n

terms, each obtained by differentiating one factor and leaving all others unchanged: y'=[f_2 f_3\cdots f_n]f_1'+[f_1 f_3\cdots f_n]f_2'+\cdots+[f_1 f_2\cdots f_{n-1}]f_n'.

What is the Difference Rule?

It follows from the Sum and Constant Multiple Rules: (f-g)'(x)=f'(x)-g'(x).

Constant Multiple Rule and Sum Rule

Product Rule

Quotient Rule

Worked Examples

Frequently Asked Questions

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