Differentiation

The gradient of a curve Paper 1

The gradient of a straight line is the same everywhere. A curve's steepness changes as you move along it, so we define the gradient of a curve at a point as the gradient of the tangent there: the straight line that just touches the curve at that point.

You cannot find the tangent's gradient from two points on it, because you only know one. Instead, take a second point on the curve, close to the first, and find the gradient of the chord joining them. Then move the second point closer and closer. The chords get closer and closer to the tangent, and their gradients approach a limiting value. That limit is the gradient of the curve.

For example, on $y=x^3$ take $P(2,8)$ and a nearby point with $x$-coordinate $2+h$. The gradient of the chord is
$$\frac{(2+h{)}^3-8}{h}=\frac{8+12h+6h^2+h^3-8}{h}=12+6h+h^2$$
With $h=1$ this is 19; with $h=0.1$ it is 12.61; with $h=0.01$ it is 12.0601. As $h$ approaches 0 the gradient approaches 12, so the gradient of $y=x^3$ at $x=2$ is 12. Only this informal idea of a limit is expected: you will not have to differentiate from first principles in general.

As the second point moves along the curve towards $P$, the chord turns towards the tangent at $P$, whose gradient is 12.

Notation for derivatives Paper 1

The function that gives the gradient at every point is the derivative. If $y=\mathrm{f}(x)$, the derivative is written $\frac{\mathrm{d}y}{\mathrm{d}x}$ or $\mathrm{f}\prime (x)$. Doing it is called differentiating.

Differentiating again gives the second derivative, $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$ or $\mathrm{f}\prime \prime (x)$. It measures how fast the gradient itself is changing, which is what makes it useful for deciding whether a stationary point is a maximum or a minimum.

Differentiating powers of x Paper 1

The pattern in the chord calculation works for every power. For any rational number $n$,
$$\frac{\mathrm{d}}{\mathrm{d}x}(x^n)=n{x}^{n-1}$$
Multiply by the power, then reduce the power by 1. Two further rules let you differentiate any sum of powers:

• a constant multiple stays: $\frac{\mathrm{d}}{\mathrm{d}x}(5x^3)=15x^2$;
• sums and differences are differentiated term by term;
• a constant on its own has derivative 0, because the graph $y=c$ is horizontal.

The rule only applies to terms written as $x^n$, so rewrite first:

• $\sqrt{x}=x^{\frac{1}{2}}$ and $\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$;
• $\frac{3}{x^2}=3x^{-2}$;
• a product such as $x^2(x+4)$ must be multiplied out, and a fraction such as $\frac{x^3+2}{x}$ split into $x^2+2x^{-1}$.

There is no rule that differentiates a product or quotient factor by factor in this paper, so expanding or splitting is essential.

The chain rule Paper 1

Expanding $(3x+1{)}^7$ before differentiating would be slow, and $\sqrt{x^2+7}$ cannot be expanded at all. Both are composite functions: a function of a function. The chain rule differentiates them directly. If $y$ is a function of $u$, and $u$ is a function of $x$, then
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}x}$$
Informally: if $y$ changes 5 times as fast as $u$, and $u$ changes 3 times as fast as $x$, then $y$ changes 15 times as fast as $x$.

For a power of a bracket the working is quick: differentiate the outside as if the bracket were a single letter, then multiply by the derivative of the inside.
$$\frac{\mathrm{d}}{\mathrm{d}x}(ax+b{)}^n=n(ax+b{)}^{n-1}\times a$$
So $\frac{\mathrm{d}}{\mathrm{d}x}(3x+1{)}^7=7(3x+1{)}^6\times 3=21(3x+1{)}^6$. The commonest error is to forget the final factor, the derivative of the inside.

The same idea gives connected rates of change, later in this lesson.

Tangents and normals Paper 1

The derivative evaluated at a point gives the gradient $m$ of the tangent there. The normal is the line through the same point perpendicular to the tangent, so its gradient is $-\frac{1}{m}$ (the two gradients multiply to $-1$).

To find either equation:

1. find the $y$-coordinate of the point if it is not given;
2. differentiate and substitute the $x$-coordinate to get $m$;
3. use $y-y_1=m(x-x_1)$ with the tangent's or the normal's gradient.

A tangent is horizontal where $\frac{\mathrm{d}y}{\mathrm{d}x}=0$, and then the normal is vertical, with equation $x=x_1$.

Increasing and decreasing functions Paper 1

A function is increasing where its graph goes up from left to right, which is where $\mathrm{f}\prime (x)>0$, and decreasing where $\mathrm{f}\prime (x)<0$. To find the set of values of $x$ for which a function is decreasing, solve the inequality $\mathrm{f}\prime (x)<0$, usually a quadratic inequality, using a sketch of $\mathrm{f}\prime (x)$.

To show that a function is increasing for all $x$, show that $\mathrm{f}\prime (x)$ is always positive. Completing the square is the usual way: if $\mathrm{f}\prime (x)=3x^2-6x+5=3(x-1{)}^2+2$, then $\mathrm{f}\prime (x)⩾2>0$ for every $x$. A quadratic $\mathrm{f}\prime (x)$ with positive $x^2$ coefficient is positive for all $x$ exactly when its discriminant is negative.

Rates of change Paper 1

$\frac{\mathrm{d}y}{\mathrm{d}x}$ is the rate of change of $y$ with respect to $x$. When the variable is time $t$, $\frac{\mathrm{d}V}{\mathrm{d}t}$ is how fast $V$ is changing per second (or per minute, and so on), and a negative value means $V$ is decreasing.

In connected rates of change, two quantities depend on each other and both change with time. The chain rule links their rates:
$$\frac{\mathrm{d}A}{\mathrm{d}t}=\frac{\mathrm{d}A}{\mathrm{d}r}\times \frac{\mathrm{d}r}{\mathrm{d}t}$$
For a circle, $A=\pi r^2$ and $\frac{\mathrm{d}A}{\mathrm{d}r}=2\pi r$. If the radius increases at 0.5 cm per second, then when $r=6$ the area increases at $2\pi (6)\times 0.5=6\pi$ cm² per second.

The method:

1. write down the rate you are given and the rate you want, as derivatives with respect to $t$;
2. find a formula linking the two variables, and differentiate it;
3. write the chain rule with the derivatives arranged so the unwanted variable cancels, substitute the particular value, and give units.

Substitute the particular value (such as $r=6$) only after differentiating.

Stationary points and their nature Paper 1

A stationary point is where the gradient is zero: $\frac{\mathrm{d}y}{\mathrm{d}x}=0$. Solve that equation for $x$, then substitute each value into the equation of the curve (not the derivative) to find $y$.

At a maximum the gradient changes from positive to zero to negative, so the gradient is decreasing and $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}<0$. At a minimum it changes from negative to zero to positive, so $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}>0$. This is the second derivative test:

• $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}<0$: maximum;
• $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}>0$: minimum.

If the question does not name a method you may instead look at the sign of $\frac{\mathrm{d}y}{\mathrm{d}x}$ just either side of the point (+ then $-$ is a maximum). Use this if the second derivative is 0, because then the test tells you nothing. Points of inflexion are not part of this syllabus.

Stationary points, together with where the curve crosses the axes, are what you need to sketch a curve. The same method solves practical problems about the greatest or least value of a quantity: write the quantity as a function of one variable, differentiate, set the derivative to zero, and confirm the nature.

The stationary points of Worked example 5. Between them the gradient is negative, so the curve falls.