Topic 7

Differentiation Rules

Most derivative questions are lost before the first rule is applied: rewrite roots as powers, spot the constants, and decide which rule the function's last operation calls for. Then apply that one rule cleanly and simplify — every recipe below is a way of making that choice automatic.

5 kinds of question. Open the one you are stuck on.

Differentiate roots, fractions and powers with the power rule

4\sqrt{x} - \dfrac{3}{x^2} + \dots

1
$Rewrite every root as a fractional power$ $\sqrt[n]{x^m} = x^{m/n}$
2
$\dfrac{c}{x^n}$ $x^2\sqrt{x} = x^{5/2}$
3
$Circle the constants — they are just numbers$ $e^{\pi}$
4
$(x^n)' = nx^{n-1}$ $\tfrac{1}{2}$
5
$Rewrite with positive exponents if asked$ $Optional: negative exponents are a correct answer unless the question says to simplify.$

Remember: Rewrite, spot constants, power rule.

See it in a problem4 worked examples

f(x) = 4\sqrt{x} - \dfrac{3}{x^2} + \dfrac{\sqrt[3]{5}}{6x^7}

1
$Rewrite every root as a fractional power$ $4\sqrt{x} = 4x^{1/2}$
2
$\dfrac{c}{x^n}$ $f(x) = 4x^{1/2} - 3x^{-2} + \dfrac{\sqrt[3]{5}}{6}x^{-7}$
3
$Circle the constants — they are just numbers$ $4$
4
$(x^n)' = nx^{n-1}$ $f'(x) = 2x^{-1/2} + 6x^{-3} - \dfrac{7\sqrt[3]{5}}{6}x^{-8}$
5
$Rewrite with positive exponents if asked$ $f'(x) = \dfrac{2}{\sqrt{x}} + \dfrac{6}{x^3} - \dfrac{7\sqrt[3]{5}}{6x^8}$

f'(x) = \dfrac{2}{\sqrt{x}} + \dfrac{6}{x^3} - \dfrac{7\sqrt[3]{5}}{6x^8}

Common mistakes4

Treating $e^{\pi}$ or $\pi^2$ as a function: there is no $x$ in them, so $\dfrac{d}{dx}e^{\pi} = 0$ , not $e^{\pi}$ or $\pi e^{\pi - 1}$ .
Differentiating a fraction by differentiating top and bottom separately. Rewrite $\dfrac{x^3 + 5}{x^2}$ as $x + 5x^{-2}$ first.
Subtracting 1 from a negative exponent the wrong way: $x^{-3} \to -3x^{-4}$ , not $-3x^{-2}$ .
Thinking $x^{-3\pi}$ needs a special rule. A constant exponent is a constant exponent: $-3\pi x^{-3\pi - 1}$ .

(2x^2 - 1)e^{x}

1
$Simplify first if you can$ $Cancel common factors, split a one-term denominator, expand a short product. Fewer rules, fewer slips.$
2
$uv$ $Ask what the last operation would be if you evaluated the function at a number.$
3
$u$ $(e^{2x})' = 2e^{2x}$
4
$u'v + uv'$ $Quotient order: "low d-high minus high d-low, over the square of what's below."$
5
$Simplify the numerator; keep the denominator factored$

Remember: Simplify, decide, list four pieces, assemble.

See it in a problem4 worked examples

f(x) = (2x^2 - 1)e^{x}

f'(x) = e^{x}(2x^2 + 4x - 1)

Common mistakes4

$(uv)' \neq u'v'$ and $\left(\dfrac{u}{v}\right)' \neq \dfrac{u'}{v'}$ .
Swapping the quotient-rule numerator to $uv' - u'v$ : that flips the sign of the whole answer.
Forgetting brackets: $-(3x - 1)(2x)$ is $-6x^2 + 2x$ , not $-6x^2 - 2x$ .
Expanding the denominator $(x^2 + 4)^2$ . Leave it squared; it is already as simple as it gets.

e^{\sin(x^2)}

1
$List the layers from the outside in$ $The outermost layer is the last thing you would press on a calculator.$
2
$Differentiate the outer layer; copy its inside unchanged$
3
$x$ $x$
4
$Tidy up: numbers and simple factors to the front$

Remember: Peel the onion: one factor per layer.

See it in a problem3 worked examples

f(x) = e^{\sin(x^2)}

1
$List the layers from the outside in$ $e^{(\cdot)}$
2
$Differentiate the outer layer; copy its inside unchanged$ $\dfrac{d}{du}e^{u} = e^{u}$
3
$x$ $\cos(x^2)$
4
$Tidy up: numbers and simple factors to the front$ $f'(x) = 2x\cos(x^2)\,e^{\sin(x^2)}$

f'(x) = 2x\cos(x^2)\,e^{\sin(x^2)}

Common mistakes3

Stopping one layer early: $\dfrac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x$ , not just $\cos(x^2)$ .
Changing the inside while differentiating the outside: $(\sin(x^2))' = \cos(x^2) \cdot 2x$ , not $\cos(2x)$ .
Reading $\sin^2 x$ as $\sin(x^2)$ . $\sin^2 x = (\sin x)^2$ , whose outer layer is the square.

3^{x^2}

b

See it in a problem4 worked examples

f(x) = 3^{x^2} + \log(5x + 1)

1
$b^{u}$ $3^{x^2}$
2
$Write each piece's outer derivative$ $3^{x^2}\log 3$
3
$u'$ $2x$
4
$Combine with the sum, product or quotient rule$ $f'(x) = 2x\,3^{x^2}\log 3 + \dfrac{5}{5x + 1}$

f'(x) = 2x\,3^{x^2}\log 3 + \dfrac{5}{5x + 1}

Common mistakes4

Writing $(3^{x})' = x\,3^{x - 1}$ . That is the power rule, which needs $x$ in the base.
Forgetting the inside on a log: $\big(\log(5x + 1)\big)' = \dfrac{5}{5x + 1}$ , not $\dfrac{1}{5x + 1}$ .
Treating $\pi^x$ and $x^{\pi}$ alike: $(\pi^x)' = \pi^x\log\pi$ , $(x^{\pi})' = \pi x^{\pi - 1}$ .
Sign slips in trig: the derivatives of $\cos$ , $\csc$ , $\cot$ all start with a minus.

f(1)

Remember: Letters first, inside first, numbers last.

See it in a problem4 worked examples

f(1) = 3

1
$Write the derivative with letters only — no numbers yet$ $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$
2
$Work out which input each factor needs$ $x = 1$
3
$Read the values from the table$ $f'(1) = -2$
4
$Substitute and compute$ $(fg)'(1) = (-2)(2) + (3)(4) = 8$

(fg)'(1) = 8

Common mistakes3

Writing $f'(g(1))$ as $f'(1)$ . The outer function is evaluated at the inside's value $g(1)$ .
Using $f'(g(1)) \cdot g(1)$ instead of $f'(g(1)) \cdot g'(1)$ .
For $\dfrac{d^2}{dx^2}f(g(x))$ , forgetting that $f'(g(x))\,g'(x)$ is a product: you need the product rule and the chain rule again.