The standard results differentiate powers of \(x\) term by term. Three further rules handle expressions built from simpler functions: a power of a bracket, a product of two functions, and a quotient of two functions.
Chain rule
| Rule | Example |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}y}{\mathrm{d}u}\times\dfrac{\mathrm{d}u}{\mathrm{d}x}\) | \(y=(x^3+2)^5\) Let \(u=x^3+2\), so \(y=u^5\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=5u^4\times 3x^2\) \(=15x^2(x^3+2)^4\) |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}[f(x)]^n=n[f(x)]^{n-1}\cdot f'(x)\) | \(y=\sqrt{4x-3}=(4x-3)^{\frac{1}{2}}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{2}(4x-3)^{-\frac{1}{2}}(4)\) \(=\dfrac{2}{\sqrt{4x-3}}\) |
The second row is the chain rule written as a shortcut for a power of a bracket: differentiate the power, keep the bracket, then multiply by the derivative of the inside.
Product rule
| Rule | Example |
|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}(uv)=u\dfrac{\mathrm{d}v}{\mathrm{d}x}+v\dfrac{\mathrm{d}u}{\mathrm{d}x}\) | \(y=x^2(2x-1)^3\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=x^2\cdot 3(2x-1)^2(2)+(2x-1)^3(2x)\) \(=2x(2x-1)^2[3x+(2x-1)]\) \(=2x(2x-1)^2(5x-1)\) |
Quotient rule
| Rule | Example |
|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{u}{v}\right)=\dfrac{v\dfrac{\mathrm{d}u}{\mathrm{d}x}-u\dfrac{\mathrm{d}v}{\mathrm{d}x}}{v^2}\) | \(y=\dfrac{x+3}{2x-5}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{(2x-5)(1)-(x+3)(2)}{(2x-5)^2}\) \(=-\dfrac{11}{(2x-5)^2}\) |
A quotient can also be rewritten as a product with a negative index, for example \(\dfrac{x+3}{2x-5}=(x+3)(2x-5)^{-1}\), and differentiated with the product rule.
In the quotient rule the numerator is \(v\dfrac{\mathrm{d}u}{\mathrm{d}x}-u\dfrac{\mathrm{d}v}{\mathrm{d}x}\) — the order matters. Swapping the two terms changes the sign of every answer.
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