Differentiation

Gradient of a curve Extended

A straight line has the same gradient everywhere. A curve does not: it can be steep in one place and almost flat in another. The gradient of a curve at a point is the gradient of the tangent at that point, the straight line that just touches the curve there without cutting across it.

To estimate a gradient from a graph:

1. Place a ruler so it touches the curve at the point and makes equal angles with the curve on either side. Draw the tangent long.
2. Choose two points on your tangent that are far apart and easy to read, ideally on grid lines.
3. Gradient $=\frac{\text{change in}y}{\text{change in}x}$, read from those two points.

A tangent that goes down from left to right has a negative gradient. Because everyone draws a slightly different tangent, an examiner accepts a range of answers for an estimate. A long tangent and points far apart keep your estimate inside that range.

The tangent at P passes through $(0,-1)$ and $(2,3)$. Its gradient is $\frac{3-(-1)}{2-0}=\frac{4}{2}=2$, so the gradient of $y=x^2$ at P is 2.

Differentiating powers of x Extended

Drawing tangents to $y=x^2$ at $x=1$, $2$ and $3$ gives gradients of $2$, $4$ and $6$. The gradient is always $2x$. Doing the same for $y=x^3$ gives $3x^2$. The pattern is the rule of differentiation: it gives a formula for the gradient at any point, so you no longer need to draw anything.

The formula for the gradient is called the derivative, written $\frac{\mathrm{d}y}{\mathrm{d}x}$ (say "d y by d x"). Cambridge expects this notation.

The rule. If $y=a{x}^n$, then $\frac{\mathrm{d}y}{\mathrm{d}x}=na{x}^{n-1}$: multiply by the power, then reduce the power by 1.

• $y=5x^3$ gives $\frac{\mathrm{d}y}{\mathrm{d}x}=15x^2$.
• $y=4x$ gives $\frac{\mathrm{d}y}{\mathrm{d}x}=4$, because $4x=4x^1$ and $x^0=1$. A straight line has a constant gradient.
• $y=7$ gives $\frac{\mathrm{d}y}{\mathrm{d}x}=0$, because $y=7$ is a horizontal line. A constant on its own always differentiates to 0.
• $y=\frac{1}{3}{x}^3$ gives $\frac{\mathrm{d}y}{\mathrm{d}x}=x^2$. The constant $a$ can be a fraction.

Sums. Differentiate term by term: $y=2x^3-5x+1$ gives $\frac{\mathrm{d}y}{\mathrm{d}x}=6x^2-5$.

Brackets first. The rule works on separate powers of $x$ only. Expand any brackets before you differentiate: $y=x(x+3)=x^2+3x$, so the derivative is $2x+3$. Differentiating each bracket separately and multiplying is wrong.

In this course the powers are positive whole numbers or zero, and there are at most three terms.

The gradient at a point Extended

Once you have $\frac{\mathrm{d}y}{\mathrm{d}x}$, the gradient at any point is found by substituting the $x$-coordinate into it. Substitute into $\frac{\mathrm{d}y}{\mathrm{d}x}$, not into $y$: putting $x$ into the equation of the curve gives the height of the curve, not its steepness.

The process also runs backwards. If you are told the gradient, set $\frac{\mathrm{d}y}{\mathrm{d}x}$ equal to that value and solve for $x$. Then substitute $x$ into the equation of the curve to find $y$, if the question asks for the point.

Stationary points Extended

A stationary point (or turning point) is a point where the gradient is zero: the tangent there is horizontal. At the top of a hill it is a maximum point; at the bottom of a valley it is a minimum point.

Method.

1. Find $\frac{\mathrm{d}y}{\mathrm{d}x}$.
2. Solve $\frac{\mathrm{d}y}{\mathrm{d}x}=0$ to find the $x$-coordinate of each stationary point.
3. Substitute each $x$ into the equation of the curve to find its $y$-coordinate.

A quadratic curve has one stationary point. A cubic curve such as $y=x^3-12x+5$ usually has two, because its derivative is a quadratic, which can have two solutions.

The graph confirms Worked example 4: a hill at $x=-2$ and a valley at $x=2$, each with a horizontal tangent.

Maximum or minimum? Extended

You may use any of three methods. Show the one you use.

1. The second derivative. Differentiate $\frac{\mathrm{d}y}{\mathrm{d}x}$ again to get $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$, then substitute the $x$-coordinate of the stationary point.

• $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}>0$: a minimum. The gradient is increasing as you pass through it, from negative to positive.
• $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}<0$: a maximum. The gradient is decreasing, from positive to negative.

2. The gradient either side. Work out $\frac{\mathrm{d}y}{\mathrm{d}x}$ at a value of $x$ just below and just above the stationary point. Positive then negative (uphill, then downhill) means a maximum. Negative then positive means a minimum. Choose values close enough that no other stationary point lies between them.

3. An accurate sketch. You know the shapes: a quadratic with a positive $x^2$ term is a U shape, so its stationary point is a minimum; with a negative $x^2$ term it is a hill, so a maximum. A cubic with a positive $x^3$ term has its maximum on the left and its minimum on the right.

You are not expected to identify points of inflection, so every stationary point you meet in this course will be a maximum or a minimum.

Using differentiation in a problem Extended

Many exam questions ask for the largest or smallest value of a quantity: the biggest area, the least cost. Write the quantity as a formula in one letter, differentiate, and set the derivative equal to zero. The letters may not be $x$ and $y$: if the area is $A$ and the length is $x$, the derivative is written $\frac{\mathrm{d}A}{\mathrm{d}x}$.

Always answer the question that was asked. If it asks for the maximum area, the value of $x$ is only a step on the way; substitute it back to find the area, and show that it really is a maximum.