Differentiating powers of xAQA A-Level Maths: Revision notes

To differentiate a power of $x$, multiply by the power and reduce the power by 1: $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^n\right)=nx^{n-1}.$ This holds for every rational $n$: positive, negative or fractional. A constant differentiates to 0, and $\frac{\mathrm{d}}{\mathrm{d}x}(x)=1$. Examples: $x^5\to5x^4$; $x^{-2}\to-2x^{-3}$; $x^{\frac12}\to\frac12x^{-\frac12}$; $7\to0$. The derivative gives the gradient of the curve at any point.

Differentiate term by term. A constant multiple stays: $\frac{\mathrm{d}}{\mathrm{d}x}(kx^n)=knx^{n-1}$. Sums and differences are differentiated separately: $y=4x^3-5x^2+7x-9\;\Rightarrow\;\frac{\mathrm{d}y}{\mathrm{d}x}=12x^2-10x+7.$ Signs carry through: the derivative of $-12x^{-3}$ is $-12\times(-3)x^{-4}=+36x^{-4}$. Notation: $y=\mathrm{f}(x)$ gives $\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{f}'(x)$.

The power rule only applies to terms of the form $kx^n$, so rewrite first using index laws:

• roots: $\sqrt{x}=x^{\frac12}$, $\sqrt[3]{x}=x^{\frac13}$
• reciprocals: $\frac{1}{x^3}=x^{-3}$ and $\frac{1}{\sqrt{x}}=x^{-\frac12}$
• brackets and quotients: expand, or divide each term by the denominator. Example: $y=5\sqrt{x}-\frac{12}{x^3}=5x^{\frac12}-12x^{-3}$, so $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac52x^{-\frac12}+36x^{-4}$. For $\mathrm{f}(x)=\frac{(x+2)^2}{\sqrt{x}}$: expand to $\frac{x^2+4x+4}{x^{\frac12}}=x^{\frac32}+4x^{\frac12}+4x^{-\frac12}$, then $\mathrm{f}'(x)=\frac32x^{\frac12}+2x^{-\frac12}-2x^{-\frac32}$.

To find the gradient at a point, substitute the $x$-value into $\frac{\mathrm{d}y}{\mathrm{d}x}$, not into $y$. For $y=4x^3-5x^2+7x-9$ at $x=-1$: $12+10+7=29$. To find where the gradient has a given value, solve $\frac{\mathrm{d}y}{\mathrm{d}x}=\text{value}$: gradient 7 gives $12x^2-10x=0$, so $x=0$ or $x=\frac56$. Check every solution against any restriction such as $x>0$. Gradient zero: $2x-\frac{16}{x^2}=0\Rightarrow x^3=8\Rightarrow x=2$.

Once you have the gradient $m$ at $x=a$, the tangent is $y-y_1=m(x-a)$ where $y_1$ is the $y$-coordinate of the point. For $y=x^2+\frac{16}{x}+1$ at $x=4$: $y=21$, $\frac{\mathrm{d}y}{\mathrm{d}x}=8-1=7$, so $y-21=7(x-4)$, giving $y=7x-7$. Set $x=0$ to find where it meets the $y$-axis: $(0,-7)$.