Differentiating powers of x and stationary pointsEdexcel A-Level Maths: Revision notes

For any rational $n$, $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{n}\right)=nx^{n-1}$, and $\frac{\mathrm{d}}{\mathrm{d}x}\left(kx^{n}\right)=knx^{n-1}$ for a constant $k$. A constant differentiates to $0$. Sums and differences are differentiated term by term. Rewrite roots and reciprocals as powers first: $\sqrt{x}=x^{1/2}$, $\frac1x=x^{-1}$, $\frac{1}{x^{2}}=x^{-2}$. Example: $y=3x^{4}-\frac{2}{x}+\sqrt{x}$ becomes $3x^{4}-2x^{-1}+x^{1/2}$, so $\frac{\mathrm{d}y}{\mathrm{d}x}=12x^{3}+2x^{-2}+\frac12x^{-1/2}$.

There is no rule yet for products or quotients, so first expand brackets or split a fraction into separate terms. Example: $y=(2x+5)(x-1)=2x^{2}+3x-5$, so $\frac{\mathrm{d}y}{\mathrm{d}x}=4x+3$. Example: $\frac{x^{2}+3x-5}{4x^{1/2}}=\frac14x^{3/2}+\frac34x^{1/2}-\frac54x^{-1/2}$, so the derivative is $\frac38x^{1/2}+\frac38x^{-1/2}+\frac58x^{-3/2}$.

The gradient of the curve at $x=a$ is $f'(a)$. The tangent there has equation $y-y_{1}=m(x-x_{1})$ with $m=f'(a)$. The normal is perpendicular to the tangent, so its gradient is $-\frac{1}{m}$. Example: $y=x^{2}+2x$ at $(1,3)$: $\frac{\mathrm{d}y}{\mathrm{d}x}=2x+2=4$. Tangent: $y-3=4(x-1)$, so $y=4x-1$. Normal: gradient $-\frac14$, so $y-3=-\frac14(x-1)$.

A stationary point is where $f'(x)=0$. Solve $f'(x)=0$ for $x$ and substitute into $y$ to get the coordinates. Use the second derivative: $f''(x)>0$ gives a minimum, and $f''(x)<0$ gives a maximum. A function is increasing where $f'(x)>0$ and decreasing where $f'(x)<0$. Example: $y=x^{3}-6x^{2}+9x+2$ has $f'(x)=3(x-1)(x-3)$, so stationary points $(1,6)$ (maximum, as $f''(1)=-6$) and $(3,2)$ (minimum, as $f''(3)=6$), and $y$ is decreasing for $1<x<3$. The same information helps with curve sketching: mark intercepts and stationary points, and note whether the curve rises or falls between them.

1. Write the quantity to optimise (volume, area, cost) in terms of one variable, using any constraint to eliminate the other.
2. Differentiate and set the derivative equal to $0$.
3. Solve (reject values that are impossible, such as negative lengths).
4. Justify the maximum or minimum with $f''$, and find the value asked for. Example: $x^{2}+4xh=1200$ gives $V=x^{2}h=300x-\frac{x^{3}}{4}$. Then $\frac{\mathrm{d}V}{\mathrm{d}x}=300-\frac34x^{2}=0$ gives $x=20$, $\frac{\mathrm{d}^{2}V}{\mathrm{d}x^{2}}=-30<0$ (maximum) and $V=4000$ cm$^{3}$.