Further differentiation

Differentiation

Standard derivatives

Composites

The chain rule gives these general results:

$6x\,\mathrm{e}^{3x^2} \qquad \frac{6x}{3x^2 + 1} \qquad 2\sec^2 2x \qquad 2\sin x\cos x = \sin 2x$

$\sin^2 x$ means $(\sin x)^2$, so use the power rule first and multiply by the derivative of $\sin x$.

The product and quotient rules

$\frac{\mathrm{d}y}{\mathrm{d}x} = x^2 \cdot 2\mathrm{e}^{2x} + \mathrm{e}^{2x}\cdot 2x = 2x\mathrm{e}^{2x}(x + 1)$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x\cdot\frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$

The quotient rule also gives these results, which are worth recognising:

$\frac{\mathrm{d}}{\mathrm{d}x}(\sec x) = \sec x\tan x \qquad \frac{\mathrm{d}}{\mathrm{d}x}(\operatorname{cosec} x) = -\operatorname{cosec} x\cot x \qquad \frac{\mathrm{d}}{\mathrm{d}x}(\cot x) = -\operatorname{cosec}^2 x$

Stationary points with these functions

Set the derivative to zero and solve. A positive factor like $\mathrm{e}^{-x}$ can never be zero, so divide it out safely.

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{-x} - x\mathrm{e}^{-x} = (1 - x)\mathrm{e}^{-x}$. Since $\mathrm{e}^{-x} > 0$, the only solution is $x = 1$, at the point $\left(1, \mathrm{e}^{-1}\right)$.

The gradient goes from positive to negative through $x = 1$, so this is a maximum.

Parametric curves

A curve can be described by writing $x$ and $y$ in terms of a third variable, the parameter $t$.

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{6t^2}{2t} = 3t$, which is 3 at $t = 1$. The point is $(2, 2)$, so the tangent is $y - 2 = 3(x - 2)$, or $y = 3x - 4$.

Implicit differentiation

When $y$ is mixed in with $x$, differentiate every term with respect to $x$. Any term in $y$ picks up a factor of $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, and a product of $x$ and $y$ needs the product rule.

$2x + x\frac{\mathrm{d}y}{\mathrm{d}x} + y + 2y\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \quad\Rightarrow\quad \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2x + y}{x + 2y}$

At $(1, 2)$ the gradient is $-\dfrac{4}{5}$.

The inverse tangent

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\tan^{-1} x\right) = \frac{1}{1 + x^2} \qquad \frac{\mathrm{d}}{\mathrm{d}x}\left(\tan^{-1} 2x\right) = \frac{2}{1 + 4x^2}$