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# AS-Level Maths | Differentiation
- URL: https://www.a-level-maths-tutor.com/as-level-maths-differentiation/
- Published: 2026-04-28T16:37:07.000Z
- Updated: 2026-07-08T16:10:54.000Z
- Author: Jaisul Naik
- Tags: Pure Maths

To differentiate $y = ax^n$: multiply the coefficient by the power and subtract one from the power, giving $\\displaystyle\\frac{dy}{dx} = nax^{n-1}$.

## Question 1

Differentiate the following expressions with respect to $x$.

**a)** $y = x^6 - 7x^2$

**b)** $y = 1 - 6x^{\\frac{5}{2}}$

**c)** $y = 2x + 8x^{-2}$

**d)** $y = x^2 - 4x^6$

**e)** $y = 5x^3 - 6x^{\\frac{3}{2}}$

**f)** $y = 9x^{-3} + 7x^{-2}$

**g)** $y = 5 - 5x^{-1}$

Show Answers 

**a)** $\\displaystyle\\frac{dy}{dx} = 6x^5 - 14x$

**b)** $\\displaystyle\\frac{dy}{dx} = -15x^{\\frac{3}{2}}$

**c)** $\\displaystyle\\frac{dy}{dx} = 2 - 16x^{-3}$

**d)** $\\displaystyle\\frac{dy}{dx} = 2x - 24x^5$

**e)** $\\displaystyle\\frac{dy}{dx} = 15x^2 - 9x^{\\frac{1}{2}}$

**f)** $\\displaystyle\\frac{dy}{dx} = -27x^{-4} - 14x^{-3}$

**g)** $\\displaystyle\\frac{dy}{dx} = 5x^{-2}$

To find the **gradient** of a curve at a specific point: differentiate to find $\\displaystyle\\frac{dy}{dx}$, then substitute the given value of $x$.

## Question 2

For each curve, find the gradient at the given value of $x$.

**a)** $y = x^3 - 5x^2 + 2$, where $x = 3$

**b)** $\\displaystyle y = 4x^2 + \\frac{3}{x}$, where $x = 1$

**c)** $y = 2x^4 - 8x$, where $x = 1$

**d)** $y = x^3 + 2x^2 - 5x + 1$, where $x = 2$

**e)** $y = 4x^2 - 3x + 7$, where $x = -2$

Show Answers 

**a)** gradient $= -3$

**b)** gradient $= 5$

**c)** gradient $= 0$

**d)** gradient $= 15$

**e)** gradient $= -19$

To find the **x-coordinate** of a stationary point: solve $\\displaystyle\\frac{dy}{dx} = 0$ for $x$.

To find the **y-coordinate** of a stationary point: substitute $x$ into the equation of the curve.

To determine the **nature** of a stationary point: substitute $x$ into $\\displaystyle\\frac{d^2y}{dx^2}$.

If $\\displaystyle\\frac{d^2y}{dx^2} < 0$ it is a maximum. If $\\displaystyle\\frac{d^2y}{dx^2} > 0$ it is a minimum.

If $\\displaystyle\\frac{d^2y}{dx^2} = 0$ check the gradient on either side.

## Question 3

For each of the following cubic equations find the coordinates of the stationary points and determine their nature.

**a)** $y = x^3 - 3x^2 - 9x + 3$

**b)** $y = x^3 + 12x^2 + 45x + 50$

**c)** $y = 2x^3 - 6x^2 + 12$

**d)** $y = 25 - 24x + 9x^2 - x^3$

Show Answers 

**a)** min$(3,\\,-24)$, max$(-1,\\,8)$

**b)** min$(-3,\\,-4)$, max$(-5,\\,0)$

**c)** min$(2,\\,4)$, max$(0,\\,12)$

**d)** min$(2,\\,5)$, max$(4,\\,9)$

## Question 4

For each of the following equations find the coordinates of the stationary points and determine their nature.

**a)** $\\displaystyle y = x + \\frac{4}{x},\\quad x \\neq 0$

**b)** $\\displaystyle y = x^2 + \\frac{16}{x},\\quad x \\neq 0$

**c)** $y = x - 4\\sqrt{x},\\quad x > 0$

**d)** $\\displaystyle y = 4x^2 + \\frac{1}{x},\\quad x \\neq 0$

Show Answers 

**a)** min$(2,\\,4)$, max$(-2,\\,-4)$

**b)** min$(2,\\,12)$

**c)** min$(4,\\,-4)$

**d)** min$\\!\\left(\\tfrac{1}{2},\\,3\\right)$

A function is **increasing** where $\\displaystyle\\frac{dy}{dx} > 0$, and **decreasing** where $\\displaystyle\\frac{dy}{dx} < 0$.

## Question 5

For each of the following equations find the range of values of $x$ for which $y$ is increasing or decreasing.

**a)** $y = 2x^3 - 3x^2 - 12x + 2$, increasing

**b)** $y = x^3 - 6x^2 + 12$, decreasing

**c)** $y = x^3 - 3x + 8$, increasing

**d)** $y = 1 - 3x^2 - x^3$, decreasing

Show Answers 

**a)** $x < -1$ or $x > 2$

**b)** $0 < x < 4$

**c)** $x < -1$ or $x > 1$

**d)** $x < -2$ or $x > 0$

To find the equation of a **tangent** at a point: use $\\displaystyle\\frac{dy}{dx}$ to find the gradient, substitute $x$ into the curve to find the $y$-coordinate, then use $y = mx + c$.

## Question 6

For each of the following curves find an equation of the tangent at the point whose $x$-coordinate is given.

**a)** $y = x^2 - 9x + 13$, where $x = 6$

**b)** $y = x^4 + x + 1$, where $x = 1$

**c)** $y = 2x^2 + 6x + 7$, where $x = -1$

**d)** $y = 2x^3 - 4x + 5$, where $x = 1$

**e)** $y = 2x^3 - 4x^2 - 3$, where $x = 2$

Show Answers 

**a)** $y = 3x - 23$

**b)** $y = 5x - 2$

**c)** $y = 2x + 5$

**d)** $y = 2x + 1$

**e)** $y = 8x - 19$

To find the equation of a **normal** at a point: use $\\displaystyle\\frac{dy}{dx}$ to find the gradient, take the negative reciprocal, substitute $x$ into the curve to find the $y$-coordinate, then use $y = mx + c$.

## Question 7

For each of the following curves find an equation of the normal at the point whose $x$-coordinate is given.

**a)** $f(x) = x^3 - 4x^2 + 1$, where $x = 2$

**b)** $f(x) = x^3 - 7x^2 + 11x$, where $x = 3$

**c)** $f(x) = 3x^4 - 7x^3 + 5$, where $x = 2$

**d)** $f(x) = \\tfrac{1}{4}x^5 - 18x + 11$, where $x = 2$

Show Answers 

**a)** $4y = x - 30$

**b)** $4y = x - 15$

**c)** $12y + x + 34 = 0$

**d)** $2y + x + 32 = 0$

## Question 8

Find the rate of change of each quantity with respect to the stated variable.

**a)** If $A = \\pi x^2 - 20x$, find the rate of change of $A$ with respect to $x$.

**b)** If $V = x - 2\\pi x^3$, find the rate of change of $V$ with respect to $x$.

**c)** If $P = at^2 - bt$, find the rate of change of $P$ with respect to $t$.

**d)** If $W = 6kh^{\\frac{1}{2}} - h$, find the rate of change of $W$ with respect to $h$.

**e)** If $N = (at + b)^2$, find the rate of change of $N$ with respect to $t$.

Show Answers 

**a)** $\\displaystyle\\frac{dA}{dx} = 2\\pi x - 20$

**b)** $\\displaystyle\\frac{dV}{dx} = 1 - 6\\pi x^2$

**c)** $\\displaystyle\\frac{dP}{dt} = 2at - b$

**d)** $\\displaystyle\\frac{dW}{dh} = 3kh^{-\\frac{1}{2}} - 1$

**e)** $\\displaystyle\\frac{dN}{dt} = 2a^2t + 2ab$

## Question 9

For each of the following equations find the coordinates of the stationary points and determine their nature.

**a)** $\\displaystyle y = \\frac{1}{x} - \\frac{1}{\\sqrt{x}},\\quad x > 0$

**b)** $\\displaystyle y = \\frac{3\\sqrt{x} - 2}{x^{\\frac{3}{2}}},\\quad x > 0$

**c)** $\\displaystyle y = \\sqrt\[3\]{x} + \\frac{27}{x},\\quad x > 0$

**d)** $\\displaystyle y = \\frac{1}{3\\sqrt{x}}\\!\\left(\\frac{2}{x} - 3\\right),\\quad x > 0$

Show Answers 

**a)** min$\\!\\left(4,\\,-\\tfrac{1}{4}\\right)$

**b)** max$(1,\\,1)$

**c)** min$(27,\\,4)$

**d)** min$\\!\\left(2,\\,-\\displaystyle\\frac{\\sqrt{2}}{3}\\right)$

The derivative from first principles is $\\displaystyle f'(x) = \\lim\_{h \\to 0} \\frac{f(x+h) - f(x)}{h}$.

## Question 10

Prove each of the following from first principles.

**a)** The derivative of $4x$ is $4$.

**b)** The derivative of $5x^2$ is $10x$.

**c)** The derivative of $3x^2$ is $6x$.

**d)** The derivative of $x^3$ is $3x^2$.

**e)** The derivative of $2x^3$ is $6x^2$.

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