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Operation Rules of Differentiation

The operation rules of differentiation are an essential tool for computing the derivatives of complicated functions. Mastering them lets you break the differentiation of a complicated function down into the differentiation of simple ones.

Sum and Difference Rule

$(u \pm v)' = u' \pm v'$

Derivation:

Let $f(x) = u(x) + v(x)$ . Starting from the definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$= \lim_{h \to 0} \frac{[u(x + h) + v(x + h)] - [u(x) + v(x)]}{h}$

$= \lim_{h \to 0} \frac{[u(x + h) - u(x)] + [v(x + h) - v(x)]}{h}$

$= \lim_{h \to 0} \frac{u(x + h) - u(x)}{h} + \lim_{h \to 0} \frac{v(x + h) - v(x)}{h}$

$= u'(x) + v'(x)$

The difference case is proved in exactly the same way, giving $(u - v)' = u' - v'$ .

Explanation: the derivative of the sum (difference) of two functions equals the sum (difference) of their derivatives.

Examples:

$(x^2 + 3x)' = (x^2)' + (3x)' = 2x + 3$
$(x^3 - x^2)' = (x^3)' - (x^2)' = 3x^2 - 2x$

Product Rule

$(uv)' = u'v + uv'$

Derivation:

Let $f(x) = u(x)v(x)$ . Starting from the definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$= \lim_{h \to 0} \frac{u(x + h)v(x + h) - u(x)v(x)}{h}$

Add and subtract $u(x)v(x + h)$ : $= \lim_{h \to 0} \frac{u(x + h)v(x + h) - u(x)v(x + h) + u(x)v(x + h) - u(x)v(x)}{h}$

$= \lim_{h \to 0} \frac{[u(x + h) - u(x)]v(x + h) + u(x)[v(x + h) - v(x)]}{h}$

$= \lim_{h \to 0} \frac{u(x + h) - u(x)}{h} \cdot \lim_{h \to 0} v(x + h) + u(x) \cdot \lim_{h \to 0} \frac{v(x + h) - v(x)}{h}$

$= u'(x)v(x) + u(x)v'(x)$

Explanation: the derivative of the product of two functions equals the derivative of the first function times the second, plus the first function times the derivative of the second.

Memory aid: $(uv)' = u'v + uv'$ (differentiate the first factor only, then the second factor only)

Examples:

$(x^2 \cdot \sin x)' = 2x \cdot \sin x + x^2 \cdot \cos x$
$(e^x \cdot x)' = e^x \cdot x + e^x \cdot 1 = e^x(x + 1)$

Quotient Rule

$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$

Derivation:

Let $f(x) = \frac{u(x)}{v(x)}$ . Starting from the definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$= \lim_{h \to 0} \frac{\frac{u(x + h)}{v(x + h)} - \frac{u(x)}{v(x)}}{h}$

Put the fractions over a common denominator: $= \lim_{h \to 0} \frac{u(x + h)v(x) - u(x)v(x + h)}{h \cdot v(x)v(x + h)}$

Add and subtract $u(x)v(x)$ : $= \lim_{h \to 0} \frac{[u(x + h) - u(x)]v(x) - u(x)[v(x + h) - v(x)]}{h \cdot v(x)v(x + h)}$

$= \lim_{h \to 0} \frac{\frac{u(x + h) - u(x)}{h} \cdot v(x) - u(x) \cdot \frac{v(x + h) - v(x)}{h}}{v(x)v(x + h)}$

$= \frac{u'(x)v(x) - u(x)v'(x)}{v^2(x)}$

Explanation: the derivative of the quotient of two functions equals the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

Memory aid: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$ (differentiate the top only minus the bottom only, then divide by the square of the bottom)

Examples:

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

Chain Rule

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Derivation:

Let $y = f(u)$ and $u = g(x)$ , so $y = f(g(x))$ .

By the definition of the derivative: $\frac{dy}{dx} = \lim_{h \to 0} \frac{f(g(x + h)) - f(g(x))}{h}$

Let $\Delta u = g(x + h) - g(x)$ , so $g(x + h) = g(x) + \Delta u$ .

As $h \to 0$ , we have $\Delta u \to 0$ (because $g(x)$ is continuous).

Hence: $\frac{dy}{dx} = \lim_{h \to 0} \frac{f(g(x) + \Delta u) - f(g(x))}{h}$

$= \lim_{h \to 0} \frac{f(g(x) + \Delta u) - f(g(x))}{\Delta u} \cdot \frac{\Delta u}{h}$

$= \lim_{\Delta u \to 0} \frac{f(g(x) + \Delta u) - f(g(x))}{\Delta u} \cdot \lim_{h \to 0} \frac{g(x + h) - g(x)}{h}$

$= f'(g(x)) \cdot g'(x) = \frac{dy}{du} \cdot \frac{du}{dx}$

Explanation: the derivative of a composite function equals the derivative of the outer function multiplied by the derivative of the inner function.

Memory aid: work from the outside in, differentiating one layer at a time.

Examples:

$(\sin(x^2))' = \cos(x^2) \cdot (x^2)' = \cos(x^2) \cdot 2x = 2x\cos(x^2)$
$(e^{x^3})' = e^{x^3} \cdot (x^3)' = e^{x^3} \cdot 3x^2 = 3x^2e^{x^3}$

Exercises

Exercise 1

Find the derivative of $f(x) = x^3 + 2x^2 - 5x + 1$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the sum and difference rule together with the power function derivative formula.

Steps:

Apply the sum and difference rule: $f'(x) = (x^3)' + (2x^2)' - (5x)' + (1)'$
Compute the derivative of each term: $(x^3)' = 3x^2$ $(2x^2)' = 2 \cdot 2x = 4x$ $(5x)' = 5$ $(1)' = 0$
Combine the results: $f'(x) = 3x^2 + 4x - 5$

Answer: $f'(x) = 3x^2 + 4x - 5$

Exercise 2

Find the derivative of $f(x) = x^2 \sin x$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the product rule.

Steps:

Let $u = x^2$ and $v = \sin x$
Compute $u'$ and $v'$ : $u' = 2x$ $v' = \cos x$
Apply the product rule: $f'(x) = u'v + uv' = 2x \cdot \sin x + x^2 \cdot \cos x$
Simplify: $f'(x) = 2x\sin x + x^2\cos x$

Answer: $f'(x) = 2x\sin x + x^2\cos x$

Exercise 3

Find the derivative of $f(x) = \frac{x^2 + 1}{x - 1}$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the quotient rule.

Steps:

Let $u = x^2 + 1$ and $v = x - 1$
Compute $u'$ and $v'$ : $u' = 2x$ $v' = 1$
Apply the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2} = \frac{2x(x-1) - (x^2+1)(1)}{(x-1)^2}$
Expand: $= \frac{2x^2 - 2x - x^2 - 1}{(x-1)^2} = \frac{x^2 - 2x - 1}{(x-1)^2}$

Answer: $f'(x) = \frac{x^2 - 2x - 1}{(x-1)^2}$

Exercise 4

Find the derivative of $f(x) = \sin(x^2 + 1)$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the chain rule.

Steps:

Let $u = x^2 + 1$ , so $f(x) = \sin u$
Compute the derivative of each layer: $\frac{df}{du} = \cos u = \cos(x^2 + 1)$ $\frac{du}{dx} = 2x$
Apply the chain rule: $f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = \cos(x^2 + 1) \cdot 2x$
Simplify: $f'(x) = 2x\cos(x^2 + 1)$

Answer: $f'(x) = 2x\cos(x^2 + 1)$

Exercise 5

Suppose $\lim\limits_{x\to1} \frac{f(x)}{\ln x}=1$ . Find the value of $f'(1)$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the property of equivalent infinitesimals together with the chain rule.

Steps:

From $\lim\limits_{x\to1} \frac{f(x)}{\ln x}=1$ we get $f(x)\sim \ln x$ as $x\to1$
As $x\to1$ , $\ln x\to0$ , so $\lim\limits_{x\to1} f(x)=0$
Let $g(x)=\frac{f(x)}{\ln x}$ , then $f(x)=g(x)\ln x$
Apply the product rule: $f'(x)=g'(x)\ln x + g(x)\cdot\frac{1}{x}$
As $x\to1$ , $\ln x\to0$ , $\frac{1}{x}\to1$ , $g(x)\to1$
Therefore $f'(1)=\lim\limits_{x\to1} f'(x)=1$

Answer: $f'(1)=1$

Exercise 6

Suppose the function $y=f(x)$ is determined by the parametric equations $\begin{cases}x=2t+|t|\\y=|t|\sin t\end{cases}$ . Find $f'(0)$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the parametric differentiation formula together with the chain rule.

Steps:

When $t\geq0$ , $x=3t, y=t\sin t$ , so $y=\frac{x}{3}\sin\frac{x}{3}$
When $t<0$ , $x=t, y=-t\sin t$ , so $y=-x\sin x$
Hence $f(x)=\begin{cases}\frac{x}{3}\sin\frac{x}{3}, & x\geq0\\-x\sin x, & x<0\end{cases}$
Compute the left derivative: $f'_-(0)=\lim\limits_{x\to0^-}\frac{-x\sin x-0}{x}=-\lim\limits_{x\to0^-}\sin x=0$
Compute the right derivative: $f'_+(0)=\lim\limits_{x\to0^+}\frac{\frac{x}{3}\sin\frac{x}{3}-0}{x}=\lim\limits_{x\to0^+}\frac{\sin\frac{x}{3}}{3}=0$
Since the left and right derivatives are equal, $f'(0)=0$

Answer: $f'(0)=0$

Exercise 7

Suppose $f(u)$ is differentiable and $z=xyf\left(\frac{y}{x}\right)$ . Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the chain rule for partial derivatives together with the product rule.

Steps:

Let $u=\frac{y}{x}$ , then $z=xyf(u)$
Differentiate with respect to $x$ : $\frac{\partial z}{\partial x}=yf(u)+xyf'(u)\cdot\frac{\partial u}{\partial x}$ $=yf(u)+xyf'(u)\cdot\left(-\frac{y}{x^2}\right)$ $=yf\left(\frac{y}{x}\right)-y^2f'\left(\frac{y}{x}\right)\frac{1}{x}$
Differentiate with respect to $y$ : $\frac{\partial z}{\partial y}=xf(u)+xyf'(u)\cdot\frac{\partial u}{\partial y}$ $=xf(u)+xyf'(u)\cdot\frac{1}{x}$ $=xf\left(\frac{y}{x}\right)+yf'\left(\frac{y}{x}\right)$

Answer: $\frac{\partial z}{\partial x}=yf\left(\frac{y}{x}\right)-y^2f'\left(\frac{y}{x}\right)\frac{1}{x}$ $\frac{\partial z}{\partial y}=xf\left(\frac{y}{x}\right)+yf'\left(\frac{y}{x}\right)$

Exercise 8

Suppose the curve $y=y(x)(x>0)$ passes through the point $(1,2)$ , and the distance from any point $P(x,y)$ on the curve to the $y$ -axis equals the intercept of the tangent line at that point on the $y$ -axis. Find $y(x)$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the tangent line equation and the notion of an intercept to set up a differential equation.

Steps:

The distance from $P(x,y)$ to the $y$ -axis is $x$
The tangent line at $P(x,y)$ is: $Y-y=y'(x)(X-x)$
Its intercept on the $y$ -axis is: when $X=0$ , $Y=y-xy'(x)$
By the hypothesis: $x=y-xy'(x)$
Rearranging: $y'(x)=\frac{y-x}{x}$
This is a first-order linear differential equation: $y'(x)-\frac{1}{x}y=-1$
Solving gives $y(x)=x(2-\ln x)$
Check $y(1)=2$ , which satisfies the condition

Answer: $y(x)=x(2-\ln x)$

Exercise 9

Given $f(x)=\int_0^x e^{\cos t}dt$ , find $f'(x)$ and $f''(x)$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the derivative formula for a function with a variable upper limit together with the chain rule.

Steps:

By the derivative formula for a function with a variable upper limit: $f'(x)=e^{\cos x}$
Differentiate $f'(x)$ once more: $f''(x)=\frac{d}{dx}(e^{\cos x})=e^{\cos x}\cdot(-\sin x)=-e^{\cos x}\sin x$
Use the chain rule: let $u=\cos x$ , then $f'(x)=e^u$ $\frac{d}{dx}(e^u)=e^u\cdot\frac{du}{dx}=e^{\cos x}\cdot(-\sin x)$

Answer: $f'(x)=e^{\cos x}$ $f''(x)=-e^{\cos x}\sin x$

Exercise 10

Given $f(x) = \int_0^x e^{t^2} \sin t\,dt$ , find $f'(x)$ and $f''(x)$ .

Reference Answer

derivative calculationdifferentiation rules

Approach: Use the derivative formula for a function with a variable upper limit together with the product rule.

Steps:

By the derivative formula for a function with a variable upper limit: $f'(x) = e^{x^2} \sin x$
Differentiate $f'(x)$ once more, using the product rule: let $u = e^{x^2}$ , $v = \sin x$

$u' = e^{x^2} \cdot 2x = 2x e^{x^2}$ $v' = \cos x$
Apply the product rule: $f''(x) = u'v + uv' = 2x e^{x^2} \cdot \sin x + e^{x^2} \cdot \cos x$ $= 2x e^{x^2} \sin x + e^{x^2} \cos x$

Answer: $f'(x) = e^{x^2} \sin x$ $f''(x) = 2x e^{x^2} \sin x + e^{x^2} \cos x$

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$u, v$	Mathematical symbol	u, v	The two differentiable functions combined by each operation rule
$u', v'$	Mathematical symbol	u prime, v prime	The derivatives of the two functions, taken one at a time in the product rule
$h$	Mathematical symbol	h	The increment in the difference quotient from which every rule on this page is derived
$(u \pm v)'$	Mathematical symbol	u plus or minus v prime	The sum and difference rule $(u \pm v)' = u' \pm v'$
$(uv)'$	Mathematical symbol	u v prime	The product rule, written $u'v + uv'$
$\left(\frac{u}{v}\right)'$	Mathematical symbol	derivative of u over v	The quotient rule, whose result carries the square $v^2$ of the denominator
$\Delta u$	Mathematical symbol	delta u	The increment of the inner function in the chain rule, with $\Delta u = g(x+h) - g(x)$
$\frac{dy}{du} \cdot \frac{du}{dx}$	Mathematical symbol	d y over d u times d u over d x	The chain rule in differential form, with $y = f(u)$ and $u = g(x)$
$\sin x$	Mathematical symbol	sine of x	The trigonometric factor in the product rule example $(x^2 \cdot \sin x)'$
$e^x$	Mathematical symbol	e to the x	The exponential factor in the product rule example $(e^x \cdot x)' = e^x(x+1)$

Basic Rules of Differentiation

Differentiation of Composite Functions

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Operation Rules of Differentiation

Sum and Difference Rule

Product Rule

Quotient Rule

Chain Rule

Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10

Summary

Symbols Used in This Article

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