C Calculus — printable coverage pack

C1 · Know that the gradient function dy/dx gives the gradient of the curve and measures the rate of change of y with respect to x

The derivative $\frac{dy}{dx}$ is the gradient function of a curve $y=f(x)$ .
At a chosen value of $x$ , substitute into $\frac{dy}{dx}$ to find the instantaneous rate of change of $y$ with respect to $x$ .
A positive derivative means $y$ is increasing locally; a negative derivative means $y$ is decreasing locally.
Do not substitute into $y$ when the question asks for a rate of change; substitute into the derivative.

Tier 1 · Easy

1. At $x=2$ , a curve has $\frac{dy}{dx}=6$ . State its instantaneous rate at this point.[1 mark]
Answer
- Rate of change $=6$
Method: $\frac{dy}{dx}$ directly gives the rate of change of $y$ with respect to $x$ . Its value at the point is $6$ .

Tier 2 · Standard

1. For $y=3x^2-5x$ , calculate $\frac{dy}{dx}$ when $x=2$ .[2 marks]
Answer
- $\dfrac{dy}{dx}=7$
Method: Differentiate: $\frac{dy}{dx}=6x-5$ . At $x=2$ , $\frac{dy}{dx}=6(2)-5=7$ .

Tier 3 · Hard

1. A curve has gradient function $\frac{dy}{dx}=3x^2-12x+5$ . Find the value of $x$ at which the rate of change is $-7$ .[3 marks]
Answer
- $x=2$
Method: Set the gradient function equal to the required rate: $3x^2-12x+5=-7$ . Then $3x^2-12x+12=0$ , so $x^2-4x+4=0$ . Since $(x-2)^2=0$ , $x=2$ .

C2 · Know that the gradient of a function is the gradient of the tangent at that point

At a point on a curve, $f'(x)$ is the gradient of the tangent at that point.
Differentiate first, then substitute the point's $x$ -coordinate into the derivative.
The tangent touches the curve locally and has the same instantaneous gradient there.
A common error is to substitute into the original function and report the $y$ -coordinate as the gradient.

Tier 1 · Easy

1. For a function $f$ , $f'(3)=-2$ . State the gradient of the tangent to $y=f(x)$ where $x=3$ .[1 mark]
Answer
- Tangent gradient $=-2$
Method: $f'(3)$ is the gradient of the tangent at $x=3$ , so the required gradient is $-2$ .

Tier 2 · Standard

1. Work out the gradient of the tangent to $y=x^2-4x+1$ at the point where $x=1$ .[2 marks]
Answer
- Tangent gradient $=-2$
Method: Differentiate to get $\frac{dy}{dx}=2x-4$ . At $x=1$ , the tangent gradient is $2(1)-4=-2$ .

Tier 3 · Hard

1. Work out the coordinates of every point on $y=x^3-3x^2+2$ where the tangent has gradient $9$ .[4 marks]
Answer
- $(-1,-2)$ and $(3,2)$
Method: $\frac{dy}{dx}=3x^2-6x$ . Set this equal to $9$ : $3x^2-6x-9=0$ , so $x^2-2x-3=0$ and $(x-3)(x+1)=0$ . Thus $x=3$ or $x=-1$ . Substitution into the curve gives $y=2$ at $x=3$ and $y=-2$ at $x=-1$ .

C3 · Differentiation of kx^n where n is an integer, and the sum of such functions

Use the power rule $\frac{d}{dx}(kx^n)=knx^{n-1}$ for integer $n$ .
Differentiate a sum term by term, keeping each sign and coefficient.
A constant differentiates to $0$ , and negative powers follow the same power rule.
A common error is to multiply by the power but forget to reduce the power by $1$ .

Tier 1 · Easy

1. Differentiate $5x^4$ with respect to $x$ .[1 mark]
Answer
- $20x^3$
Method: Multiply the coefficient by the power and reduce the power by $1$ : $\frac{d}{dx}(5x^4)=5\times4x^3=20x^3$ .

Tier 2 · Standard

1. Given $y=4x^3-3x^2+7x-9$ , find $\frac{dy}{dx}$ .[3 marks]
Answer
- $\frac{dy}{dx}=12x^2-6x+7$
Method: Differentiate term by term: $4x^3$ gives $12x^2$ , $-3x^2$ gives $-6x$ , $7x$ gives $7$ , and the constant $-9$ gives $0$ . Therefore $\frac{dy}{dx}=12x^2-6x+7$ .

Tier 3 · Hard

1. For $y=2x^5-\frac{3}{x^2}+4x$ , find $\frac{dy}{dx}$ and hence find its value at $x=1$ .[4 marks]
Answer
- $\frac{dy}{dx}=10x^4+\frac{6}{x^3}+4$
- $\dfrac{dy}{dx}=20$ when $x=1$
Method: Write $-\frac{3}{x^2}$ as $-3x^{-2}$ . Differentiating gives $10x^4+6x^{-3}+4=10x^4+\frac{6}{x^3}+4$ . At $x=1$ , the value is $10+6+4=20$ .

C4 · The equation of a tangent and normal at any point on a curve

Differentiate the curve and substitute the point's $x$ -coordinate to find the tangent gradient.
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal when both gradients are defined.
Use the point on the curve in $y-y_1=m(x-x_1)$ for either line.
If only an $x$ -coordinate is given, first substitute into the curve to find the corresponding $y$ -coordinate.

Tier 1 · Easy

1. The curve $y=x^2$ passes through $(2,4)$ . State the gradients of the tangent and the normal at this point.[2 marks]
Answer
- Tangent gradient $=4$
- Normal gradient $=-\frac{1}{4}$
Method: $\frac{dy}{dx}=2x$ , so at $x=2$ the tangent gradient is $4$ . The normal gradient is the negative reciprocal, $-\frac{1}{4}$ .

Tier 2 · Standard

1. Work out the equation of the tangent to $y=x^2+3x-1$ at the point where $x=1$ .[3 marks]
Answer
- $y=5x-2$
Method: At $x=1$ , $y=1+3-1=3$ , so the point is $(1,3)$ . Also $\frac{dy}{dx}=2x+3$ , giving tangent gradient $5$ . Thus $y-3=5(x-1)$ , so $y=5x-2$ .

Tier 3 · Hard

1. The normal to $y=x^3-x$ at the point where $x=2$ meets the $x$ -axis at $R$ . Find the coordinates of $R$ .[4 marks]
Answer
- $R=(68,0)$
Method: At $x=2$ , $y=2^3-2=6$ , so the point is $(2,6)$ . Since $\frac{dy}{dx}=3x^2-1$ , the tangent gradient is $11$ and the normal gradient is $-\frac{1}{11}$ . The normal is $y-6=-\frac{1}{11}(x-2)$ . At the $x$ -axis $y=0$ , so $-6=-\frac{1}{11}(x-2)$ , giving $66=x-2$ and $x=68$ .

C5 · Increasing and decreasing functions

A differentiable function is increasing where $\frac{dy}{dx}>0$ and decreasing where $\frac{dy}{dx}<0$ .
Find boundary values by solving $\frac{dy}{dx}=0$ , then test the sign of the derivative in each interval.
Values where the derivative is $0$ are stationary candidates or interval boundaries; determine increasing and decreasing intervals from the derivative's sign on either side.
A common error is to inspect whether $y$ is positive or negative instead of checking the sign of the derivative.

Tier 1 · Easy

1. A function has derivative $f'(x)=2x-6$ . State the values of $x$ for which the function is increasing.[2 marks]
Answer
- $x>3$
Method: The function is increasing when $f'(x)>0$ . Solve $2x-6>0$ : $2x>6$ , so $x>3$ .

Tier 2 · Standard

1. Work out where $y=x^2-8x+1$ is decreasing and where it is increasing.[3 marks]
Answer
- Decreasing for $x<4$
- Increasing for $x>4$
Method: $\frac{dy}{dx}=2x-8=2(x-4)$ . This derivative is negative when $x<4$ and positive when $x>4$ . Therefore the curve decreases before $x=4$ and increases after $x=4$ .

Tier 3 · Hard

1. Work out the intervals on which $f(x)=x^3-3x^2-9x+4$ is increasing and the interval on which it is decreasing.[4 marks]
Answer
- Increasing for $x<-1$ and $x>3$
- Decreasing for $-1<x<3$
Method: $f'(x)=3x^2-6x-9=3(x-3)(x+1)$ . The derivative is $0$ at $x=-1$ and $x=3$ . It is positive outside these roots and negative between them, so $f$ increases for $x<-1$ and $x>3$ , and decreases for $-1<x<3$ .

C6 · Understand and use the notation d2y/dx2; know that it measures the rate of change of the gradient function

$\frac{d^2y}{dx^2}$ is the derivative of $\frac{dy}{dx}$ , so it measures how the gradient changes as $x$ changes.
Differentiate the original function twice, simplifying negative or fractional powers before or after differentiating as convenient.
At a point, $\frac{d^2y}{dx^2}>0$ means the gradient is increasing and $\frac{d^2y}{dx^2}<0$ means the gradient is decreasing.
A common error is to square $\frac{dy}{dx}$ ; the superscript $2$ records a second differentiation, not a square.

Tier 1 · Easy

1. For $y=3x^4-5x^2+7$ , work out $\frac{d^2y}{dx^2}$ .[2 marks]
Answer
- $\frac{d^2y}{dx^2}=36x^2-10$
Method: Differentiate once: $\frac{dy}{dx}=12x^3-10x$ . Differentiate again: $\frac{d^2y}{dx^2}=36x^2-10$ .

Tier 2 · Standard

1. A curve has $\frac{dy}{dx}=6x^2-4x$ . Work out the value of $x$ at which the gradient is changing at a rate of $20$ , given $x>0$ .[3 marks]
Answer
- $x=2$
Method: The rate of change of the gradient is $\frac{d^2y}{dx^2}$ . Differentiate again: $\frac{d^2y}{dx^2}=12x-4$ . Set $12x-4=20$ , so $12x=24$ and $x=2$ (consistent with $x>0$ ).

Tier 3 · Hard

1. For the curve $y=x^4-4x^3+2x^2$ , work out the ranges of $x$ for which the gradient is increasing. Give exact values.[4 marks]
Answer
- $x<1-\frac{\sqrt{6}}{3}$ or $x>1+\frac{\sqrt{6}}{3}$
Method: First $\frac{dy}{dx}=4x^3-12x^2+4x$ , so $\frac{d^2y}{dx^2}=12x^2-24x+4=4(3x^2-6x+1)$ . The gradient is increasing when $3x^2-6x+1>0$ . The boundary values are $x=\frac{6\pm\sqrt{36-12}}{6}=1\pm\frac{\sqrt{6}}{3}$ . The quadratic opens upwards, so it is positive outside the two roots.

C7 · Use of differentiation to find maxima and minima points on a curve

At a stationary point, $\frac{dy}{dx}=0$ ; solve this equation and substitute each $x$ -value into the original curve to obtain coordinates.
If $\frac{d^2y}{dx^2}<0$ at a stationary point it is a local maximum; if $\frac{d^2y}{dx^2}>0$ it is a local minimum.
A change of gradient from positive to negative also proves a maximum, while negative to positive proves a minimum.
Always give the full coordinates when a question asks for points, and state their nature rather than stopping at the stationary $x$ -values.

Tier 1 · Easy

1. The curve $y=x^3-3x^2+2$ has a stationary point where $x=2$ . Use $\frac{d^2y}{dx^2}$ to determine its nature.[2 marks]
Answer
- The stationary point is a minimum.
Method: $\frac{dy}{dx}=3x^2-6x$ , so $\frac{d^2y}{dx^2}=6x-6$ . At $x=2$ , $\frac{d^2y}{dx^2}=6>0$ , so the stationary point is a minimum.

Tier 2 · Standard

1. The curve $y=x^3-6x^2+9x+1$ has stationary points where $x=1$ and $x=3$ . Work out their coordinates and determine their nature.[3 marks]
Answer
- $(1,5)$ is a maximum.
- $(3,1)$ is a minimum.
Method: Substitution gives $y(1)=1-6+9+1=5$ and $y(3)=27-54+27+1=1$ . Also $\frac{d^2y}{dx^2}=6x-12$ . This is $-6$ at $x=1$ , giving a maximum, and $6$ at $x=3$ , giving a minimum.

Tier 3 · Hard

1. Work out the coordinates and nature of every stationary point of $y=2x^4-16x^2+3$ .[4 marks]
Answer
- $(-2,-29)$ is a minimum.
- $(0,3)$ is a maximum.
- $(2,-29)$ is a minimum.
Method: $\frac{dy}{dx}=8x^3-32x=8x(x-2)(x+2)$ , so $x=-2,0,2$ . Substitution gives $y=-29,3,-29$ respectively. Since $\frac{d^2y}{dx^2}=24x^2-32$ , its values are $64,-32,64$ . Positive values give minima at $(-2,-29)$ and $(2,-29)$ ; the negative value gives a maximum at $(0,3)$ .

C8 · Using calculus to find maxima and minima in simple problems

Define the quantity to be optimised as a function of one variable, using any length, perimeter or other constraint to remove extra variables.
Differentiate the resulting function and solve $\frac{dA}{dx}=0$ , or the corresponding derivative for the quantity in the question.
Check that the stationary value is the required maximum or minimum, using $\frac{d^2A}{dx^2}$ , a sign change or the shape of a quadratic.
Return to the requested quantity: a stationary input value is not the final answer when the question asks for a maximum area or volume.
Reject values that are impossible in the context, such as negative lengths or values outside the stated domain.

Tier 1 · Easy

1. A quantity is modelled by $P=x(14-x)$ for $0<x<14$ . Find the greatest possible value of $P$ by differentiating.[2 marks]
Answer
- The maximum value of $P$ is $49$ .
Method: $P=14x-x^2$ , so $\frac{dP}{dx}=14-2x$ . Setting this to zero gives $x=7$ . Then $P(7)=7(14-7)=49$ ; the negative $x^2$ coefficient confirms a maximum.

Tier 2 · Standard

1. A rectangle has side lengths $(x+2)$ cm and $(10-x)$ cm, where $0<x<10$ . Use calculus to work out its maximum area.[3 marks]
Answer
- The maximum area is $36\text{ cm}^2$ .
Method: $A=(x+2)(10-x)=-x^2+8x+20$ . Hence $\frac{dA}{dx}=-2x+8$ , so the stationary value is at $x=4$ . The second derivative is $-2<0$ , so this is a maximum. $A(4)=6\times6=36\text{ cm}^2$ .

Tier 3 · Hard

1. The upper corners of a rectangle lie on $y=12-x^2$ at $(x,y)$ and $(-x,y)$ , where $x>0$ . The lower corners lie on the $x$ -axis. Use calculus to work out the maximum area of the rectangle.[4 marks]
Answer
- The maximum area is $32$ square units.
Method: The width is $2x$ and the height is $12-x^2$ , so $A=2x(12-x^2)=24x-2x^3$ . Then $\frac{dA}{dx}=24-6x^2$ . Setting this to zero and using $x>0$ gives $x=2$ . Also $\frac{d^2A}{dx^2}=-12x=-24<0$ , so this is a maximum. $A(2)=4(12-4)=32$ .

C9 · Sketch/interpret a curve with known maximum and minimum points

A local maximum is a turning point where the curve changes from increasing to decreasing; a local minimum changes from decreasing to increasing.
On a sketch, plot and label the known turning points and intercepts before joining them with a smooth curve.
Use the sign of the gradient to decide which sections rise or fall as $x$ increases.
A sketch shows the required shape and key features; it does not need an exact scale, but it must not introduce extra turning points or intercepts.

Tier 1 · Easy

1. A smooth curve has a local maximum at $x=-2$ and a local minimum at $x=1$ , with no other stationary points. State where the curve is decreasing.[1 mark]
Answer
- The curve is decreasing for $-2<x<1$ .
Method: After a local maximum the curve falls, and it continues to fall until the local minimum. Therefore it is decreasing between the two stationary $x$ -values.

Tier 2 · Standard

1. Sketch the cubic $y=2x^3-3x^2-12x+2$ , which has local maximum $L=(-1,9)$ and local minimum $M=(2,-18)$ . Label $L$ , $M$ and the $y$ -intercept $N$ .[3 marks]
Answer
- A smooth cubic rising to $L=(-1,9)$ , falling through $N=(0,2)$ to $M=(2,-18)$ , then rising again.
Method: At $x=0$ , $y=2$ , so $N=(0,2)$ . Plot $L$ , $M$ and $N$ . The positive-leading cubic comes from the lower left and rises to $L$ , decreases through $N$ to $M$ , then rises towards the upper right. Join the points smoothly with horizontal tangents at $L$ and $M$ .

Tier 3 · Hard

1. A continuous smooth curve $y=f(x)$ has roots $x=-4$ , $x=1$ and $x=5$ . It has a local maximum at $(-2,5)$ and a local minimum at $(3,-4)$ , with no other turning points. Sketch a possible curve, labelling all five given points.[4 marks]
Answer
- A smooth curve crossing at $(-4,0)$ , rising to $(-2,5)$ , crossing at $(1,0)$ , falling to $(3,-4)$ , then crossing at $(5,0)$ and rising.
Method: Place the three roots on the $x$ -axis and the maximum and minimum at their stated coordinates. The order in $x$ is $-4,-2,1,3,5$ . Draw one smooth curve that crosses each root, turns at the maximum and minimum only, and has horizontal tangents at those turning points.