Differentiation Rules

Power Rule & Basic Rules

Constant Rule

\[ \frac{d}{dx}[c] = 0 \]

The derivative of any constant is zero.

Power Rule

\[ \frac{d}{dx}[x^n] = nx^{n-1} \]

Bring the exponent down and subtract 1 from it. Works for all real \(n\).

Constant Multiple Rule

\[ \frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x) \]

Constants factor out of derivatives.

Sum / Difference Rule

\[ \frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x) \]

Differentiate term by term.

Example 1

Find \(f'(x)\) if \(f(x) = 5x^3 - 7x^2 + 4x - 9\).

\(f'(x) = 5(3x^2) - 7(2x) + 4(1) - 0 = 15x^2 - 14x + 4\)

Example 2

Find \(\dfrac{d}{dx}\left[3\sqrt{x} + \dfrac{2}{x^2}\right]\).

Rewrite: \(3x^{1/2} + 2x^{-2}\)

\(= 3 \cdot \tfrac{1}{2}x^{-1/2} + 2(-2)x^{-3} = \dfrac{3}{2\sqrt{x}} - \dfrac{4}{x^3}\)

Practice Problems

1. \(f(x) = x^7\)

\(f'(x) = 7x^6\)

2. \(g(x) = 4x^5 - 3x^2 + 8\)

\(g'(x) = 20x^4 - 6x\)

3. \(h(x) = \sqrt[3]{x^2} = x^{2/3}\)

\(h'(x) = \dfrac{2}{3}x^{-1/3} = \dfrac{2}{3\sqrt[3]{x}}\)

4. \(y = \dfrac{5}{x^4}\)

Rewrite as \(5x^{-4}\), so \(y' = -20x^{-5} = -\dfrac{20}{x^5}\)

Product & Quotient Rules

Product Rule

\[ \frac{d}{dx}[f(x)\cdot g(x)] = f'(x)g(x) + f(x)g'(x) \]

Memory: "first times derivative of second, plus second times derivative of first."

Quotient Rule

\[ \frac{d}{dx}\!\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]

Memory: "low d-high minus high d-low, over low squared." \(\left(\frac{L\cdot DH - H\cdot DL}{L^2}\right)\)

Product Example

Differentiate \(y = (3x^2+1)(x^4-5x)\).

Let \(f = 3x^2+1\), \(g = x^4-5x\).

\(f' = 6x\), \(g' = 4x^3-5\)

\(y' = 6x(x^4-5x) + (3x^2+1)(4x^3-5)\)

\(= 6x^5-30x^2 + 12x^5-15x^2+4x^3-5 = 18x^5+4x^3-45x^2-5\)

Quotient Example

Differentiate \(y = \dfrac{x^2+3}{2x-1}\).

\(y' = \dfrac{2x(2x-1) - (x^2+3)(2)}{(2x-1)^2} = \dfrac{4x^2-2x-2x^2-6}{(2x-1)^2} = \dfrac{2x^2-2x-6}{(2x-1)^2}\)

Practice Problems

1. \(y = (2x-3)(x^2+4)\)

\(y' = 2(x^2+4) + (2x-3)(2x) = 2x^2+8+4x^2-6x = 6x^2-6x+8\)

2. \(y = \dfrac{3x}{x^2+1}\)

\(y' = \dfrac{3(x^2+1)-3x(2x)}{(x^2+1)^2} = \dfrac{3-3x^2}{(x^2+1)^2}\)

3. \(y = x^3 \cdot e^x\)

\(y' = 3x^2 e^x + x^3 e^x = x^2 e^x(3+x)\)

Chain Rule

\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

Differentiate the outer function (leaving the inner alone), then multiply by the derivative of the inner function.

Example 1 — Power of a function

\(y = (3x^2+5)^4\)

Outer: \(u^4 \Rightarrow 4u^3\); Inner: \(3x^2+5 \Rightarrow 6x\)

\(y' = 4(3x^2+5)^3 \cdot 6x = 24x(3x^2+5)^3\)

Example 2 — Trig with chain

\(y = \sin(5x^2)\)

\(y' = \cos(5x^2) \cdot 10x = 10x\cos(5x^2)\)

Example 3 — Exponential with chain

\(y = e^{x^3}\)

\(y' = e^{x^3} \cdot 3x^2 = 3x^2 e^{x^3}\)

Example 4 — Nested chain

\(y = \sqrt{\sin(2x)} = [\sin(2x)]^{1/2}\)

\(y' = \tfrac{1}{2}[\sin(2x)]^{-1/2} \cdot \cos(2x) \cdot 2 = \dfrac{\cos(2x)}{\sqrt{\sin(2x)}}\)

Practice Problems

1. \(y = (x^3-2x)^5\)

\(y' = 5(x^3-2x)^4(3x^2-2)\)

2. \(y = \cos(4x)\)

\(y' = -\sin(4x)\cdot 4 = -4\sin(4x)\)

3. \(y = \ln(x^2+1)\)

\(y' = \dfrac{1}{x^2+1}\cdot 2x = \dfrac{2x}{x^2+1}\)

4. \(y = e^{4x-7}\)

\(y' = e^{4x-7} \cdot 4 = 4e^{4x-7}\)

Trig, Exponential & Logarithmic Derivatives

Trigonometric Derivatives

\[\frac{d}{dx}[\sin x]=\cos x \qquad \frac{d}{dx}[\cos x]=-\sin x\] \[\frac{d}{dx}[\tan x]=\sec^2 x \qquad \frac{d}{dx}[\cot x]=-\csc^2 x\] \[\frac{d}{dx}[\sec x]=\sec x\tan x \qquad \frac{d}{dx}[\csc x]=-\csc x\cot x\]

Exponential & Logarithmic Derivatives

\[\frac{d}{dx}[e^x]=e^x \qquad \frac{d}{dx}[a^x]=a^x\ln a\] \[\frac{d}{dx}[\ln x]=\frac{1}{x} \qquad \frac{d}{dx}[\log_a x]=\frac{1}{x\ln a}\]

Example — Combined

Differentiate \(y = e^x \sin x\).

Product rule: \(y' = e^x\sin x + e^x\cos x = e^x(\sin x + \cos x)\)

Example — Log with chain

Differentiate \(y = \ln(\cos x)\).

\(y' = \dfrac{1}{\cos x}\cdot(-\sin x) = -\tan x\)

Practice Problems

1. \(y = \tan(3x)\)

\(y' = \sec^2(3x)\cdot 3 = 3\sec^2(3x)\)

2. \(y = 5^x\)

\(y' = 5^x \ln 5\)

3. \(y = \ln(x^3)\)

\(y' = \dfrac{3x^2}{x^3} = \dfrac{3}{x}\) (or use log property: \(3\ln x \Rightarrow 3/x\))

Inverse Trigonometric Derivatives

All Six Inverse Trig Derivatives

\[\frac{d}{dx}[\sin^{-1}x] = \frac{1}{\sqrt{1-x^2}}\] \[\frac{d}{dx}[\cos^{-1}x] = \frac{-1}{\sqrt{1-x^2}}\] \[\frac{d}{dx}[\tan^{-1}x] = \frac{1}{1+x^2}\] \[\frac{d}{dx}[\cot^{-1}x] = \frac{-1}{1+x^2}\] \[\frac{d}{dx}[\sec^{-1}x] = \frac{1}{|x|\sqrt{x^2-1}}\] \[\frac{d}{dx}[\csc^{-1}x] = \frac{-1}{|x|\sqrt{x^2-1}}\]

Pattern: cos⁻¹, cot⁻¹, and csc⁻¹ are the negatives of sin⁻¹, tan⁻¹, and sec⁻¹ respectively.

Example 1

Differentiate \(y = \sin^{-1}(3x)\).

Chain rule: \(y' = \dfrac{1}{\sqrt{1-(3x)^2}}\cdot 3 = \dfrac{3}{\sqrt{1-9x^2}}\)

Example 2

Differentiate \(y = \tan^{-1}(x^2)\).

Chain rule: \(y' = \dfrac{1}{1+(x^2)^2}\cdot 2x = \dfrac{2x}{1+x^4}\)

Example 3

Differentiate \(y = x\cdot\sin^{-1}x\).

Product rule: \(y' = 1\cdot\sin^{-1}x + x\cdot\dfrac{1}{\sqrt{1-x^2}} = \sin^{-1}x + \dfrac{x}{\sqrt{1-x^2}}\)

Practice Problems

1. \(y = \cos^{-1}(2x)\)

\(y' = \dfrac{-1}{\sqrt{1-(2x)^2}}\cdot 2 = \dfrac{-2}{\sqrt{1-4x^2}}\)

2. \(y = \tan^{-1}(5x)\)

\(y' = \dfrac{5}{1+25x^2}\)

3. \(y = \sec^{-1}(x^3)\)

\(y' = \dfrac{1}{|x^3|\sqrt{x^6-1}}\cdot 3x^2 = \dfrac{3x^2}{|x^3|\sqrt{x^6-1}}\)

4. \(y = \cot^{-1}(e^x)\)

\(y' = \dfrac{-1}{1+(e^x)^2}\cdot e^x = \dfrac{-e^x}{1+e^{2x}}\)

Quick Reference — All Derivative Rules

Basic Rules

Constant

\(\frac{d}{dx}[c]=0\)

Power Rule

\(\frac{d}{dx}[x^n]=nx^{n-1}\)

Constant Multiple

\(\frac{d}{dx}[cf]=c\cdot f'\)

Sum / Difference

\(\frac{d}{dx}[f\pm g]=f'\pm g'\)

Product Rule

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

Quotient Rule

\(\left(\tfrac{f}{g}\right)'=\tfrac{f'g-fg'}{g^2}\)

Chain Rule

\(\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)\)

Exponential & Logarithmic

Natural Exp

\(\frac{d}{dx}[e^x]=e^x\)

General Exp

\(\frac{d}{dx}[a^x]=a^x\ln a\)

Natural Log

\(\frac{d}{dx}[\ln x]=\frac{1}{x}\)

General Log

\(\frac{d}{dx}[\log_a x]=\frac{1}{x\ln a}\)

Trigonometric

sin

\(\frac{d}{dx}[\sin x]=\cos x\)

cos

\(\frac{d}{dx}[\cos x]=-\sin x\)

tan

\(\frac{d}{dx}[\tan x]=\sec^2 x\)

cot

\(\frac{d}{dx}[\cot x]=-\csc^2 x\)

sec

\(\frac{d}{dx}[\sec x]=\sec x\tan x\)

csc

\(\frac{d}{dx}[\csc x]=-\csc x\cot x\)

Inverse Trigonometric

sin⁻¹

\(\dfrac{1}{\sqrt{1-x^2}}\)

cos⁻¹

\(\dfrac{-1}{\sqrt{1-x^2}}\)

tan⁻¹

\(\dfrac{1}{1+x^2}\)

cot⁻¹

\(\dfrac{-1}{1+x^2}\)

sec⁻¹

\(\dfrac{1}{|x|\sqrt{x^2-1}}\)

csc⁻¹

\(\dfrac{-1}{|x|\sqrt{x^2-1}}\)

🎯 Differentiation Quiz

8 questions — click your answer to check instantly.

1. \(\dfrac{d}{dx}[x^8] = \)

2. \(\dfrac{d}{dx}[e^x \cdot \cos x] = \)

3. Chain rule: \(\dfrac{d}{dx}[(x^2+1)^6] = \)

4. \(\dfrac{d}{dx}[\tan x] = \)

5. Quotient rule: \(\dfrac{d}{dx}\!\left[\dfrac{x^2}{x+1}\right] = \)

6. \(\dfrac{d}{dx}[\ln(x^2+5)] = \)

7. \(\dfrac{d}{dx}[\sin^{-1}(x)] = \)

8. \(\dfrac{d}{dx}[\tan^{-1}(2x)] = \)