Gradient of a curve and differentiation from first principlesAQA A-Level Maths: Revision notes

A curve has a different gradient at each point. The gradient at a point is defined as the gradient of the tangent there, the straight line that touches the curve and has the same direction. The gradient function (derivative) of $y=\mathrm{f}(x)$ gives the gradient at a general point $(x,y)$. It is written $\mathrm{f}'(x)$ or $\frac{\mathrm{d}y}{\mathrm{d}x}$. Because gradient is the rate at which $y$ changes with $x$, the derivative is a rate of change: if $h$ is height and $t$ is time, $\frac{\mathrm{d}h}{\mathrm{d}t}$ is the velocity.

Take $P(x,\mathrm{f}(x))$ and a nearby point $Q$ with $x$-coordinate $x+h$. The gradient of the chord $PQ$ is $\frac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}.$ As $h$ gets smaller, $Q$ approaches $P$ and the chord approaches the tangent. The gradient of the tangent is the limit as $h\to0$: $\mathrm{f}'(x)=\lim_{h\to0}\frac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}.$ This is differentiation from first principles. Example: $P(3,9)$ on $y=x^2$ gives chord gradient $\frac{(3+h)^2-9}{h}=6+h$, which tends to 6 as $h\to0$.

For $\mathrm{f}(x)=x^2$: $\frac{(x+h)^2-x^2}{h}=\frac{2xh+h^2}{h}=2x+h\to2x$. For $\mathrm{f}(x)=x^3$: $(x+h)^3=x^3+3x^2h+3xh^2+h^3$, so $\frac{(x+h)^3-x^3}{h}=3x^2+3xh+h^2\to3x^2$. For $x^4$ the binomial expansion gives $4x^3$. The pattern is $\frac{\mathrm{d}}{\mathrm{d}x}(x^n)=nx^{n-1}$. For a function such as $3x^2-5x$, apply the definition to the whole expression: $\frac{6xh+3h^2-5h}{h}=6x+3h-5\to6x-5$. Always write 'as $h\to0$' in your working.

Once you know $\mathrm{f}'(x)$, the gradient at $x=a$ is $\mathrm{f}'(a)$ and the tangent there is $y-\mathrm{f}(a)=\mathrm{f}'(a)(x-a)$. For $\mathrm{f}(x)=3x^2-5x$ at $x=2$: $\mathrm{f}(2)=2$, $\mathrm{f}'(2)=7$, so $y=7x-12$. To find where the gradient has a given value, solve $\mathrm{f}'(x)=\text{value}$. Interpretation: the mean rate of change between two points is the gradient of a chord, while the instantaneous rate of change is the gradient of the tangent. For $h=12t-t^3$ between $t=1$ and $t=1.1$ the mean velocity is 8.69 m s⁻¹ but the instantaneous velocity at $t=1$ is 9 m s⁻¹.

Differentiating $\frac{\mathrm{d}y}{\mathrm{d}x}$ again gives the second derivative, written $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ or $\mathrm{f}''(x)$. It is the rate of change of the gradient. Example: $y=x^3$ has $\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=6x$. For $h=12t-t^3$, $\frac{\mathrm{d}h}{\mathrm{d}t}=12-3t^2$ is the velocity and $\frac{\mathrm{d}^2h}{\mathrm{d}t^2}=-6t$ is the rate of change of velocity (acceleration). A negative $\mathrm{f}''(x)$ means the gradient is decreasing.

To sketch $y=\mathrm{f}'(x)$ from the graph of $y=\mathrm{f}(x)$, read the gradient at several points. Where the curve has a minimum or maximum the gradient is 0, so $\mathrm{f}'$ crosses or touches the $x$-axis. Where the curve is increasing, $\mathrm{f}'(x)>0$ (graph above the axis); where decreasing, $\mathrm{f}'(x)<0$. Where the curve is steepest, $|\mathrm{f}'(x)|$ is largest. A curve with a minimum at $x=1$ has a gradient function that is negative for $x<1$, zero at $x=1$ and positive for $x>1$. A quadratic has a straight-line gradient function; a cubic has a quadratic one.