2.2 Differentiation Rules
Shortcut rules that replace the limit definition — powers, products, quotients, trigonometric, exponential and inverse trigonometric functions.
Finding derivatives from the definition can be lengthy and, for some functions, quite challenging. We found $\dfrac{d}{dx}\left(\sqrt{x}\right)=\dfrac{1}{2\sqrt{x}}$ only after multiplying by a conjugate; doing $\dfrac{d}{dx}\left(\sqrt[3]{x}\right)$ that way is worse. In this section we develop rules that bypass the process entirely.
The Basic Rules
The functions $f(x)=c$ and $g(x)=x^n$ (with $n$ a positive integer) are the building blocks from which all polynomials and rational functions are constructed, so we start there.
📐 Theorem 2.2 — The Constant Rule
Let $c$ be a constant. If $f(x)=c$, then $f'(x)=0$; equivalently $\dfrac{d}{dx}(c)=0$.
From the definition, both $f(x)=c$ and $f(x+h)=c$, so the difference quotient is $\dfrac{c-c}{h}=0$ for every $h\neq0$. A constant function is a horizontal line, and a horizontal line has slope $0$.
✏️ Example 1 — Differentiating $x^3$
Find $\dfrac{d}{dx}\left(x^3\right)$.
💡 Show solution▶
Step 1 Expand $(x+h)^3=x^3+3x^2h+3xh^2+h^3$ in the difference quotient:
$$\frac{d}{dx}\left(x^3\right)=\lim_{h\to0}\frac{(x+h)^3-x^3}{h}=\lim_{h\to0}\frac{3x^2h+3xh^2+h^3}{h}.$$
Step 2 Factor out $h$ and cancel:
$$=\lim_{h\to0}\frac{h\left(3x^2+3xh+h^2\right)}{h}=\lim_{h\to0}\left(3x^2+3xh+h^2\right)=3x^2.$$
The pattern The power $3$ became the coefficient, and the power dropped by one. That is the power rule.
📐 Theorem 2.3 — The Power Rule
Let $n$ be a positive integer. If $f(x)=x^n$, then
$$f'(x)=nx^{n-1},\qquad\text{that is}\qquad \frac{d}{dx}\left(x^n\right)=nx^{n-1}.$$
⚠️ The power rule needs the variable in the base
The rule applies to $x^n$, where the base varies and the exponent is fixed. It does not apply to a constant raised to a variable power such as $f(x)=3^x$ — that needs Theorem 2.12 further down.
✏️ Examples 2 and 3 — Applying the constant and power rules
Find the derivative of $f(x)=8$, and the derivative of $f(x)=x^{10}$.
💡 Show solutions▶
Example 2 One step with the constant rule: $\dfrac{d}{dx}(8)=0$.
Example 3 Power rule with $n=10$: $f'(x)=10x^{10-1}=10x^9$.
📐 Theorem 2.4 — Sum, Difference and Constant Multiple Rules
Let $f$ and $g$ be differentiable and $k$ a constant.
Sum rule$\dfrac{d}{dx}\big(f(x)+g(x)\big)=\dfrac{d}{dx}\big(f(x)\big)+\dfrac{d}{dx}\big(g(x)\big)$
Difference rule$\dfrac{d}{dx}\big(f(x)-g(x)\big)=\dfrac{d}{dx}\big(f(x)\big)-\dfrac{d}{dx}\big(g(x)\big)$
Constant multiple rule$\dfrac{d}{dx}\big(kf(x)\big)=k\,\dfrac{d}{dx}\big(f(x)\big)$
✏️ Example 4 — Applying the basic derivative rules
Find the derivative of $f(x)=2x^5+7$.
💡 Show solution▶
Step 1 Sum rule first:
$$f'(x)=\frac{d}{dx}\left(2x^5+7\right)=\frac{d}{dx}\left(2x^5\right)+\frac{d}{dx}(7).$$
Step 2 Constant multiple rule on the first term, constant rule on the second:
$$=2\frac{d}{dx}\left(x^5\right)+0.$$
Step 3 Power rule: $=2\left(5x^4\right)=10x^4$.
The Product Rule
It is tempting to assume the derivative of a product is the product of the derivatives, but that pattern fails: for $f(x)=x\cdot x=x^2$ the derivative is $2x$, not $\dfrac{d}{dx}(x)\cdot\dfrac{d}{dx}(x)=1\cdot1=1$.
📐 Theorem 2.5 — The Product Rule
Let $f(x)$ and $g(x)$ be differentiable. Then
$$\frac{d}{dx}\big(f(x)g(x)\big)=\frac{d}{dx}\big(f(x)\big)\cdot g(x)+f(x)\cdot\frac{d}{dx}\big(g(x)\big),$$
that is, $\big(f(x)g(x)\big)'=f'(x)g(x)+f(x)g'(x)$: the derivative of the first times the second, plus the first times the derivative of the second.
✏️ Example 5 — Applying the product rule to binomials
For $j(x)=\left(x^2+2\right)\left(3x^3-5x\right)$, find $j'(x)$ by applying the product rule.
💡 Show solution▶
Step 1 Apply the rule:
$$j'(x)=\left(x^2+2\right)'\left(3x^3-5x\right)+\left(x^2+2\right)\left(3x^3-5x\right)'.$$
Step 2 Differentiate each factor:
$$=(2x)\left(3x^3-5x\right)+\left(x^2+2\right)\left(9x^2-5\right).$$
Step 3 Expand and collect:
$$=6x^4-10x^2+9x^4-5x^2+18x^2-10=15x^4+3x^2-10.$$
Check Multiplying out first gives $j(x)=3x^5+x^3-10x$, whose derivative is $15x^4+3x^2-10$ — the same answer.
The Quotient Rule
📐 Theorem 2.6 — The Quotient Rule
Let $f(x)$ and $g(x)$ be differentiable. Then
$$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{\dfrac{d}{dx}\big(f(x)\big)\cdot g(x)-f(x)\cdot\dfrac{d}{dx}\big(g(x)\big)}{\big(g(x)\big)^2}.$$
Again the naive pattern fails: $\dfrac{d}{dx}\left(x^2\right)=2x$, whereas $\dfrac{\frac{d}{dx}\left(x^3\right)}{\frac{d}{dx}(x)}=3x^2$.
✏️ Example 6 — Applying the quotient rule
Use the quotient rule to find the derivative of $k(x)=\dfrac{5x^2}{4x+3}$.
💡 Show solution▶
Step 1 Apply the rule:
$$k'(x)=\frac{\left(5x^2\right)'(4x+3)-\left(5x^2\right)(4x+3)'}{(4x+3)^2}=\frac{10x(4x+3)-\left(5x^2\right)(4)}{(4x+3)^2}.$$
Step 2 Expand the numerator: $40x^2+30x-20x^2=20x^2+30x$, so
$$k'(x)=\frac{20x^2+30x}{(4x+3)^2}.$$
📐 Theorem 2.7 — Extended Power Rule
If $k$ is a negative integer, then $\dfrac{d}{dx}\left(x^k\right)=kx^{k-1}$.
So $\dfrac{d}{dx}\left(\dfrac{1}{x}\right)=\dfrac{d}{dx}\left(x^{-1}\right)=-x^{-2}=-\dfrac{1}{x^2}$, which matches the answer we obtained from the definition in Section 2.1.
🧭 Extending the product rule to three factors
Let $f$, $g$ and $h$ be differentiable. Then
$$\big(f(x)g(x)h(x)\big)'=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x).$$
Differentiate one factor at a time and leave the others alone.
Combining Differentiation Rules
It seldom happens that just one rule is needed. A good rule of thumb: apply the differentiation rules in the reverse of the order in which you would evaluate the function.
✏️ Example 7 — Combining the quotient rule and the product rule
For $h(x)=\dfrac{2x^3k(x)}{3x+2}$, find $h'(x)$.
💡 Show solution▶
Step 1 The outermost operation is a quotient, so start there:
$$h'(x)=\frac{\left(2x^3k(x)\right)'(3x+2)-\left(2x^3k(x)\right)(3x+2)'}{(3x+2)^2}.$$
Step 2 The numerator of the original contains a product, so apply the product rule inside:
$$=\frac{\left(6x^2k(x)+2x^3k'(x)\right)(3x+2)-\left(2x^3k(x)\right)(3)}{(3x+2)^2}.$$
Step 3 Expand and collect:
$$=\frac{18x^3k(x)+12x^2k(x)+6x^4k'(x)+4x^3k'(x)-6x^3k(x)}{(3x+2)^2}=\frac{12x^3k(x)+12x^2k(x)+6x^4k'(x)+4x^3k'(x)}{(3x+2)^2}.$$
Note $k$ is an unspecified differentiable function, so $k$ and $k'$ simply stay in the answer.
✏️ Example 8 — Finding a velocity
The position of an object on a coordinate axis at time $t$ is $s(t)=\dfrac{t}{t^2+1}$. What is the initial velocity of the object?
💡 Show solution▶
Step 1 The initial velocity is $v(0)=s'(0)$, so differentiate with the quotient rule:
$$s'(t)=\frac{1\left(t^2+1\right)-t(2t)}{\left(t^2+1\right)^2}=\frac{1-t^2}{\left(t^2+1\right)^2}.$$
Step 2 Evaluate at $t=0$:
$$v(0)=s'(0)=\frac{1-0}{\left(0+1\right)^2}=1.$$
Higher-Order Derivatives
The derivative of a function is itself a function, so it too can be differentiated. The derivative of position is velocity; the derivative of velocity is acceleration. The function obtained by differentiating the derivative is the second derivative, and continuing gives the third, fourth and higher-order derivatives. The notation:
$$f''(x),\ f'''(x),\ f^{(4)}(x),\dots,f^{(n)}(x)\qquad y'',\ y''',\ y^{(4)},\dots,y^{(n)}\qquad \frac{d^2y}{dx^2},\ \frac{d^3y}{dx^3},\ \frac{d^4y}{dx^4},\dots,\frac{d^ny}{dx^n}$$The notation $\dfrac{d^2y}{dx^2}$ can be read as a compact way of writing $\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$, and similarly $\dfrac{d}{dx}\left(\dfrac{d^2y}{dx^2}\right)=\dfrac{d^3y}{dx^3}$.
✏️ Example 9 — Finding a second derivative
For $f(x)=2x^2-3x+1$, find $f''(x)$.
💡 Show solution▶
Step 1 Differentiate once: $f'(x)=4x-3$.
Step 2 Differentiate again: $f''(x)=4$.
Derivatives of Trigonometric Functions
One of the most important types of motion in physics is simple harmonic motion, described with sine and cosine functions. Being able to differentiate them lets us find the velocity and acceleration of such motion.
📐 Theorem 2.8 — The derivatives of $\sin x$ and $\cos x$
$$\frac{d}{dx}(\sin x)=\cos x\qquad\qquad \frac{d}{dx}(\cos x)=-\sin x$$
✏️ Example 10 — Differentiating a function containing $\sin x$
Find the derivative of $f(x)=5x^3\sin x$.
💡 Show solution▶
Step 1 This is a product, so apply the product rule:
$$f'(x)=\left(5x^3\right)'\sin x+5x^3(\sin x)'=15x^2\sin x+5x^3\cos x.$$
✏️ Example 11 — A function containing $\cos x$
Find the derivative of $g(x)=\dfrac{\cos x}{4x^2}$.
💡 Show solution▶
Step 1 Quotient rule:
$$g'(x)=\frac{(\cos x)'\left(4x^2\right)-(\cos x)\left(4x^2\right)'}{\left(4x^2\right)^2}=\frac{(-\sin x)\left(4x^2\right)-(\cos x)(8x)}{16x^4}.$$
Step 2 Simplify by cancelling $4x$:
$$=\frac{-4x^2\sin x-8x\cos x}{16x^4}=\frac{-x\sin x-2\cos x}{4x^3}.$$
📐 Theorem 2.9 — Derivatives of $\tan x$, $\cot x$, $\sec x$ and $\csc x$
tangent$\dfrac{d}{dx}(\tan x)=\sec^2x$
cotangent$\dfrac{d}{dx}(\cot x)=-\csc^2x$
secant$\dfrac{d}{dx}(\sec x)=\sec x\tan x$
cosecant$\dfrac{d}{dx}(\csc x)=-\csc x\cot x$
✏️ Example 12 — Derivative of trigonometric functions
Find the derivative of $f(x)=\csc x+x\tan x$.
💡 Show solution▶
Step 1 Sum rule first:
$$f'(x)=\frac{d}{dx}(\csc x)+\frac{d}{dx}(x\tan x).$$
Step 2 Product rule on the second term:
$$f'(x)=-\csc x\cot x+(1)(\tan x)+(x)\left(\sec^2x\right)=-\csc x\cot x+\tan x+x\sec^2x.$$
Exponential and Logarithmic Functions
The function $f(x)=b^x$ with $b>0$, $b\neq1$ is the general exponential function, and its inverse $g(x)=\log_b x$ is the general logarithmic function. In particular $E(x)=e^x$ is the natural exponential function and $L(x)=\ln x$ is the natural logarithmic function.
📐 Theorems 2.10–2.12 — Exponential and logarithmic derivatives
Theorem 2.10If $E(x)=e^x$ then $E'(x)=e^x$
Theorem 2.11If $x>0$ and $y=\ln x$ then $\dfrac{dy}{dx}=\dfrac1x$
Theorem 2.12 (log)If $y=\log_b x$ then $\dfrac{dy}{dx}=\dfrac{1}{x\ln b}$
Theorem 2.12 (exp)If $y=b^x$ then $\dfrac{dy}{dx}=b^x\ln b$
The natural exponential is the one function that is its own derivative — the reason $e$ matters at all.
✏️ Example 13 — Applying derivative formulas
Find the derivative of $h(x)=\dfrac{3^x}{3^x+2}$.
💡 Show solution▶
Step 1 Quotient rule, with $\dfrac{d}{dx}\left(3^x\right)=3^x\ln3$:
$$h'(x)=\frac{3^x\ln3\left(3^x+2\right)-3^x\left(3^x\ln3\right)}{\left(3^x+2\right)^2}.$$
Step 2 The $3^{2x}\ln3$ terms cancel, leaving
$$h'(x)=\frac{2\cdot3^x\ln3}{\left(3^x+2\right)^2}.$$
Note Do not drop the factor $\ln 3$: the cancellation leaves $2\cdot3^x\ln3$ in the numerator.
Derivatives of Inverse Trigonometric Functions
These derivatives are surprising: they are algebraic functions. Until now derivatives of algebraic functions were algebraic and derivatives of trigonometric functions were trigonometric. As with $\ln x$, whose derivative $\tfrac1x$ is algebraic, the derivative here is not of the same type as the original function.
📐 Theorem 2.13 — Derivatives of inverse trigonometric functions
arcsine$\dfrac{d}{dx}\left(\sin^{-1}x\right)=\dfrac{1}{\sqrt{1-x^2}}$
arccosine$\dfrac{d}{dx}\left(\cos^{-1}x\right)=\dfrac{-1}{\sqrt{1-x^2}}$
arctangent$\dfrac{d}{dx}\left(\tan^{-1}x\right)=\dfrac{1}{1+x^2}$
arccotangent$\dfrac{d}{dx}\left(\cot^{-1}x\right)=\dfrac{-1}{1+x^2}$
arcsecant$\dfrac{d}{dx}\left(\sec^{-1}x\right)=\dfrac{1}{|x|\sqrt{x^2-1}}$
arccosecant$\dfrac{d}{dx}\left(\csc^{-1}x\right)=\dfrac{-1}{|x|\sqrt{x^2-1}}$
✏️ Example 14 — Applying formulas to an inverse sine function
Find the derivative of $h(x)=x^2\sin^{-1}x$.
💡 Show solution▶
Step 1 Product rule, with Theorem 2.13 for the second factor:
$$h'(x)=2x\sin^{-1}x+x^2\cdot\frac{1}{\sqrt{1-x^2}}.$$
Exercises
I. Find $f'(x)$ for each function
I
- 1 $f(x)=5x^3-x+1$
- 2 $f(x)=4x^2-7x$
- 3 $f(x)=8x^4+9x^2-1$
- 4 $f(x)=x^4+\dfrac2x$
- 5 $f(x)=3x\left(18x^4+\dfrac{13}{x+1}\right)$
- 6 $f(x)=x^2\left(\dfrac{2}{x^2}+\dfrac{5}{x^3}\right)$
- 7 $f(x)=(x+2)\left(2x^2-3\right)$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
1 Power, constant multiple and difference rules: $f'(x)=15x^2-1$.
2 $f'(x)=8x-7$.
3 $f'(x)=32x^3+18x$.
4 Write $\dfrac2x=2x^{-1}$ and use the extended power rule:
$$f'(x)=4x^3-\frac{2}{x^2}.$$
5 Expand first: $f(x)=54x^5+\dfrac{39x}{x+1}$. Then the power rule and the quotient rule give
$$f'(x)=270x^4+\frac{39(x+1)-39x}{(x+1)^2}=270x^4+\frac{39}{(x+1)^2}.$$
6 Simplify before differentiating: $f(x)=2+\dfrac5x$, so
$$f'(x)=-\frac{5}{x^2}.$$
Multiplying out first turns a product rule into a one-line power rule.
7 Product rule: $(1)\left(2x^2-3\right)+(x+2)(4x)=2x^2-3+4x^2+8x=6x^2+8x-3$.
I
Continued.
- 8 $f(x)=x^2\cot x$
- 9 $f(x)=\sin x\tan x$
- 10 $f(x)=(x+\cos x)(1-\sin x)$
- 11 $f(x)=x^2e^x$
- 12 $f(x)=x\csc^{-1}x$
- 13 $f(x)=3x^5\cot^{-1}x$
- 14 $f(x)=\dfrac{10^x}{\ln 10}$
- 15 $f(x)=x^\pi\cdot\pi^x$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
8 Product rule with $(\cot x)'=-\csc^2x$:
$$f'(x)=2x\cot x-x^2\csc^2x.$$
9 Product rule: $\cos x\tan x+\sin x\sec^2x=\sin x+\sin x\sec^2x=\sin x\left(1+\sec^2x\right)$.
10 Product rule:
$$f'(x)=(1-\sin x)(1-\sin x)+(x+\cos x)(-\cos x)=(1-\sin x)^2-(x+\cos x)\cos x.$$
11 Product rule with $\left(e^x\right)'=e^x$:
$$f'(x)=2xe^x+x^2e^x=x(x+2)e^x.$$
12 Product rule with Theorem 2.13:
$$f'(x)=\csc^{-1}x+x\cdot\frac{-1}{|x|\sqrt{x^2-1}}=\csc^{-1}x-\frac{x}{|x|\sqrt{x^2-1}},$$
which for $x>0$ is $\csc^{-1}x-\dfrac{1}{\sqrt{x^2-1}}$.
13 Product rule with $\left(\cot^{-1}x\right)'=\dfrac{-1}{1+x^2}$:
$$f'(x)=15x^4\cot^{-1}x-\frac{3x^5}{1+x^2}.$$
14 $\ln 10$ is a constant, and $\left(10^x\right)'=10^x\ln10$, so
$$f'(x)=\frac{10^x\ln10}{\ln10}=10^x.$$
15 A power function times an exponential function — do not confuse the two rules:
$$f'(x)=\pi x^{\pi-1}\pi^x+x^\pi\pi^x\ln\pi=x^{\pi-1}\pi^x\left(\pi+x\ln\pi\right).$$
I
Continued.
- 16 $f(x)=\dfrac{x^3+2x^2-4}{3}$
- 17 $f(x)=\dfrac{4x^3-2x+1}{x^2}$
- 18 $f(x)=\dfrac{x^2+4}{x^2-4}$
- 19 $f(x)=\dfrac{x+9}{x^2-7x+1}$
- 20 $f(x)=\dfrac{\sec x}{x}$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
16 The $3$ is a constant divisor, so no quotient rule is needed:
$$f'(x)=\frac{3x^2+4x}{3}=x^2+\frac{4x}{3}.$$
17 Split the fraction first: $f(x)=4x-\dfrac2x+\dfrac{1}{x^2}=4x-2x^{-1}+x^{-2}$, so
$$f'(x)=4+\frac{2}{x^2}-\frac{2}{x^3}.$$
18 Quotient rule:
$$f'(x)=\frac{2x\left(x^2-4\right)-\left(x^2+4\right)2x}{\left(x^2-4\right)^2}=\frac{-16x}{\left(x^2-4\right)^2}.$$
19 Quotient rule:
$$f'(x)=\frac{\left(x^2-7x+1\right)-(x+9)(2x-7)}{\left(x^2-7x+1\right)^2}=\frac{-x^2-18x+64}{\left(x^2-7x+1\right)^2}.$$
20 Quotient rule with $(\sec x)'=\sec x\tan x$:
$$f'(x)=\frac{x\sec x\tan x-\sec x}{x^2}=\frac{\sec x\left(x\tan x-1\right)}{x^2}.$$
I
Continued — the last six.
- 21 $f(x)=\dfrac{\tan x}{1-\sec x}$
- 22 $f(x)=\dfrac{1-\cot x}{1+\cot x}$
- 23 $f(x)=\dfrac{1+\cos x}{1-\cos x}$
- 24 $f(x)=\dfrac{\sec^{-1}x}{x^2+5}$
- 25 $f(x)=\dfrac{3x+5}{\cot^{-1}x}$
- 26 $f(x)=\dfrac{\log_2 x}{\cos^{-1}x}$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
21 Quotient rule:
$$f'(x)=\frac{\sec^2x(1-\sec x)+\tan x\sec x\tan x}{(1-\sec x)^2}.$$
Using $\tan^2x=\sec^2x-1$ the numerator becomes $\sec^2x-\sec^3x+\sec^3x-\sec x=\sec x(\sec x-1)$, so
$$f'(x)=\frac{\sec x(\sec x-1)}{(1-\sec x)^2}=\frac{-\sec x}{1-\sec x}=\frac{1}{1-\cos x}.$$
22 Quotient rule with $(\cot x)'=-\csc^2 x$, so $(1-\cot x)'=+\csc^2x$ and $(1+\cot x)'=-\csc^2x$:
$$f'(x)=\frac{\csc^2x(1+\cot x)-(1-\cot x)\left(-\csc^2x\right)}{(1+\cot x)^2}=\frac{\csc^2x\left[(1+\cot x)+(1-\cot x)\right]}{(1+\cot x)^2}=\frac{2\csc^2x}{(1+\cot x)^2}.$$
23 Quotient rule:
$$f'(x)=\frac{-\sin x(1-\cos x)-(1+\cos x)\sin x}{(1-\cos x)^2}=\frac{-2\sin x}{(1-\cos x)^2}.$$
24 Quotient rule with $\left(\sec^{-1}x\right)'=\dfrac{1}{|x|\sqrt{x^2-1}}$:
$$f'(x)=\frac{\dfrac{x^2+5}{|x|\sqrt{x^2-1}}-2x\sec^{-1}x}{\left(x^2+5\right)^2}.$$
25 Quotient rule with $\left(\cot^{-1}x\right)'=\dfrac{-1}{1+x^2}$:
$$f'(x)=\frac{3\cot^{-1}x+\dfrac{3x+5}{1+x^2}}{\left(\cot^{-1}x\right)^2}.$$
26 Quotient rule with $\left(\log_2x\right)'=\dfrac{1}{x\ln2}$ and $\left(\cos^{-1}x\right)'=\dfrac{-1}{\sqrt{1-x^2}}$:
$$f'(x)=\frac{\dfrac{\cos^{-1}x}{x\ln2}+\dfrac{\log_2x}{\sqrt{1-x^2}}}{\left(\cos^{-1}x\right)^2}.$$
II. Higher-order derivatives
II
Find the requested higher-order derivative.
- 1 $\dfrac{d^3y}{dx^3}$ of $y=3\cos x$
- 2 $\dfrac{d^2y}{dx^2}$ of $y=3\sin x+x^2\cos x$
- 3 $\dfrac{d^4y}{dx^4}$ of $y=5\cos x$
Derivatives stacked on one set of axes
$x$—
$y$—
$y'$—
$y''$—
$y'''$—
💡 Show solutions▶
1 Differentiate three times:
$$y'=-3\sin x,\qquad y''=-3\cos x,\qquad \frac{d^3y}{dx^3}=3\sin x.$$
2 First derivative, using the product rule on the second term:
$$y'=3\cos x+2x\cos x-x^2\sin x.$$
Differentiating again:
$$\frac{d^2y}{dx^2}=-3\sin x+2\cos x-2x\sin x-2x\sin x-x^2\cos x=-3\sin x+2\cos x-4x\sin x-x^2\cos x.$$
3 Cosine returns to itself after four differentiations:
$$y'=-5\sin x,\quad y''=-5\cos x,\quad y'''=5\sin x,\quad \frac{d^4y}{dx^4}=5\cos x.$$
III–VI. More practice with the rules
III
Find $f'(x)$. Rewrite radicals and reciprocals as powers first, and simplify before reaching for the product or quotient rule when you can.
- 1 $f(x)=6\sqrt{x}-\dfrac{4}{x^3}+2$
- 2 $f(x)=\sqrt[3]{x^2}+\dfrac{1}{\sqrt{x}}$
- 3 $f(x)=\left(\sqrt{x}+1\right)\left(x^2-3\right)$
- 4 $f(x)=\dfrac{2x-1}{x^2+1}$
- 5 $f(x)=\dfrac{x^2+3x}{\sqrt{x}}$
- 6 $f(x)=\dfrac{1+x}{1-x}$
Check your answer against the graph
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
💡 Show solutions▶
1 Write $f(x)=6x^{1/2}-4x^{-3}+2$. The power rule also holds for fractional exponents such as $\tfrac12$, so
$$f'(x)=6\cdot\tfrac12x^{-1/2}-4(-3)x^{-4}+0=\frac{3}{\sqrt{x}}+\frac{12}{x^4}.$$
2 Write $f(x)=x^{2/3}+x^{-1/2}$. Then
$$f'(x)=\tfrac23x^{-1/3}-\tfrac12x^{-3/2}=\frac{2}{3\sqrt[3]{x}}-\frac{1}{2x\sqrt{x}}.$$
3 Product rule with $\left(\sqrt{x}+1\right)'=\dfrac{1}{2\sqrt{x}}$:
$$f'(x)=\frac{x^2-3}{2\sqrt{x}}+\left(\sqrt{x}+1\right)(2x)=\tfrac12x^{3/2}-\frac{3}{2\sqrt{x}}+2x^{3/2}+2x=\tfrac52x^{3/2}+2x-\frac{3}{2\sqrt{x}}.$$
4 Quotient rule:
$$f'(x)=\frac{2\left(x^2+1\right)-(2x-1)(2x)}{\left(x^2+1\right)^2}=\frac{2x^2+2-4x^2+2x}{\left(x^2+1\right)^2}=\frac{-2x^2+2x+2}{\left(x^2+1\right)^2}.$$
5 Divide first — no quotient rule needed: $f(x)=x^{3/2}+3x^{1/2}$, so
$$f'(x)=\tfrac32x^{1/2}+\tfrac32x^{-1/2}=\frac32\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)=\frac{3(x+1)}{2\sqrt{x}}.$$
6 Quotient rule:
$$f'(x)=\frac{(1)(1-x)-(1+x)(-1)}{(1-x)^2}=\frac{1-x+1+x}{(1-x)^2}=\frac{2}{(1-x)^2}.$$
The derivative is positive wherever it is defined, so $f$ increases on each side of $x=1$.
IV
Find the requested higher-order derivative.
- 1 $f''(x)$ for $f(x)=x^5-3x^3+2x$
- 2 $\dfrac{d^3y}{dx^3}$ for $y=xe^x$
- 3 $\dfrac{d^3y}{dx^3}$ for $y=\ln x$
- 4 $\dfrac{d^2y}{dx^2}$ for $y=x\sin x$
Derivatives stacked on one set of axes
$x$—
$y$—
$y'$—
$y''$—
$y'''$—
💡 Show solutions▶
1 Power rule twice:
$$f'(x)=5x^4-9x^2+2,\qquad f''(x)=20x^3-18x.$$
2 Product rule each time; every differentiation adds one more $e^x$:
$$y'=e^x+xe^x=(x+1)e^x,\qquad y''=e^x+(x+1)e^x=(x+2)e^x,\qquad \frac{d^3y}{dx^3}=(x+3)e^x.$$
3 Start with Theorem 2.11, then use the extended power rule:
$$y'=x^{-1},\qquad y''=-x^{-2},\qquad \frac{d^3y}{dx^3}=2x^{-3}=\frac{2}{x^3}\quad(x>0).$$
4 Product rule for $y'$, then again for the $x\cos x$ term:
$$y'=\sin x+x\cos x,\qquad \frac{d^2y}{dx^2}=\cos x+\left(\cos x-x\sin x\right)=2\cos x-x\sin x.$$
V
Find $f'(x)$ for each trigonometric, exponential, logarithmic or inverse trigonometric function.
- 1 $f(x)=\sec x\tan x$
- 2 $f(x)=\dfrac{e^x}{1+x}$
- 3 $f(x)=x\ln x-x$
- 4 $f(x)=5^x+\log_5x$
- 5 $f(x)=\left(1+x^2\right)\tan^{-1}x$
- 6 Conceptual. $f(x)=\sin^{-1}x+\cos^{-1}x$ on $(-1,1)$. Find $f'(x)$ and explain what the answer says about $f$.
💡 Show solutions▶
1 Product rule with $(\sec x)'=\sec x\tan x$ and $(\tan x)'=\sec^2x$:
$$f'(x)=\sec x\tan x\cdot\tan x+\sec x\cdot\sec^2x=\sec x\tan^2x+\sec^3x.$$
Using $\tan^2x=\sec^2x-1$ this is also $2\sec^3x-\sec x$.
2 Quotient rule with $\left(e^x\right)'=e^x$:
$$f'(x)=\frac{e^x(1+x)-e^x(1)}{(1+x)^2}=\frac{xe^x}{(1+x)^2}.$$
3 Product rule on $x\ln x$, then subtract the derivative of $x$:
$$f'(x)=(1)\ln x+x\cdot\frac1x-1=\ln x+1-1=\ln x.$$
So $x\ln x-x$ is a function whose derivative is $\ln x$ — a fact that returns in integration.
4 Theorem 2.12, both halves:
$$f'(x)=5^x\ln5+\frac{1}{x\ln5}.$$
5 Product rule with $\left(\tan^{-1}x\right)'=\dfrac{1}{1+x^2}$:
$$f'(x)=2x\tan^{-1}x+\left(1+x^2\right)\cdot\frac{1}{1+x^2}=2x\tan^{-1}x+1.$$
6 By Theorem 2.13,
$$f'(x)=\frac{1}{\sqrt{1-x^2}}+\frac{-1}{\sqrt{1-x^2}}=0.$$
A function with zero derivative on an interval never changes, so $f$ is constant. Evaluating at $x=0$: $f(0)=0+\tfrac{\pi}{2}$, hence $\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ for every $x$ in $[-1,1]$.
VI
Applied rates of change.
- 1 The cost, in OMR, of producing $x$ units of a product is $C(x)=2000+12x-0.01x^2+0.00002x^3$. Find the marginal cost $C'(x)$ and evaluate $C'(500)$. Interpret the result.
- 2 A bacteria culture contains $P(t)=1000\cdot2^t$ cells after $t$ hours. How fast is the population growing when $t=3$?
- 3 The concentration of a drug in the bloodstream $t$ hours after an injection is $C(t)=\dfrac{5t}{t^2+4}$ mg/L. Find $C'(t)$, then the rate of change at $t=1$ and at $t=3$. What does the change of sign tell you?
💡 Show solutions▶
1 Power, constant multiple and sum rules:
$$C'(x)=12-0.02x+0.00006x^2.$$
At $x=500$: $C'(500)=12-10+0.00006(250\,000)=12-10+15=17$. When 500 units are being produced, each additional unit costs about 17 OMR — the marginal cost approximates the cost of the 501st unit.
2 Theorem 2.12 with the constant multiple rule: $P'(t)=1000\cdot2^t\ln2$. At $t=3$:
$$P'(3)=8000\ln2\approx5545\ \text{cells per hour}.$$
3 Quotient rule:
$$C'(t)=\frac{5\left(t^2+4\right)-5t(2t)}{\left(t^2+4\right)^2}=\frac{5\left(4-t^2\right)}{\left(t^2+4\right)^2}.$$
At $t=1$: $C'(1)=\dfrac{15}{25}=0.6$ mg/L per hour. At $t=3$: $C'(3)=\dfrac{-25}{169}\approx-0.148$ mg/L per hour.
The concentration is rising at first and falling later; the switch happens where $C'(t)=0$, at $t=2$ hours, when the concentration peaks at $C(2)=\tfrac{10}{8}=1.25$ mg/L.
🔬 Research–Teaching NexusResearch-led · optional, about 30 minutes
Research question. How fast does the error of a difference quotient shrink as $h$ gets smaller, and which formula is better?
Why it matters. Numerical derivatives are used in weather forecasting, machine learning and engineering simulation; choosing the right formula and step size is a standard research question.
What to do.
- Take $f(x)=x^3$ at $x=2$; the power rule gives $f'(2)=12$.
- In a spreadsheet, compute the forward difference $\frac{f(2+h)-f(2)}{h}$ and the central difference $\frac{f(2+h)-f(2-h)}{2h}$ for $h=0.1, 0.01, 0.001, 0.0001$.
- Record the error of each. When $h$ is divided by 10, how does each error change?
- Repeat for a product such as $x^2\sin x$ at $x=1$ and check against the product rule.
Tools: Spreadsheet or calculator
Hand in: A half-page note: your error table, the pattern you found, and which formula you recommend.
Going further. Anywhere: try $h=10^{-10}$ and $10^{-12}$ and explain why the error grows again.
Instructor note: Students should see the forward error falls about 10 times per step ($\propto h$) and the central error about 100 times ($\propto h^2$).
🧪 Self-check quiz
0/5 correct
Q1. $\dfrac{d}{dx}\left(3^x\right)$ is:
Q2. For $f(x)=x^2\sin x$, $f'(x)$ is:
Q3. For $f(x)=x\tan^{-1}x$, the value $f'(1)$ is:
Q4. If $y=5\cos x$, then $\dfrac{d^4y}{dx^4}$ is:
Q5. $\dfrac{d}{dx}\left(\dfrac{x}{x+1}\right)$ is: