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Differentiation Calculator

Find derivatives step by step. Enter a polynomial expression and see the derivative with full working. Covers power rule, chain rule, product rule, quotient rule and partial differentiation — ideal for A-Level and university maths.

Reviewed by Mustafa Bilgic, Maths Specialist A-Level Aligned Free to Use

Find the Derivative (dy/dx)

How to enter expressions: Use x^2 for x², 3x^4 for 3x&sup4;, sin(x), cos(x), e^x, ln(x). Separate polynomial terms with + or -. Example: 3x^2 + 5x + 2
Supported: ax^n, sin(x), cos(x), tan(x), e^x, ln(x), constants. Polynomial terms separated by + or -
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Complete Derivatives Reference Table

The most comprehensive quick-reference table of standard derivatives for A-Level and university maths. Bookmark this page for instant lookup during revision.

Function f(x) Derivative f'(x) Rule / Notes
POWER FUNCTIONS
c (constant)0Constant rule
x1Power rule
2xPower rule
3x²Power rule
x^nnx^(n-1)Power rule — general
ax^nanx^(n-1)Constant multiple
1/x = x^(-1)-1/x²Power rule
√x = x^(1/2)1/(2√x)Power rule
x^(1/n)(1/n)x^(1/n - 1)Power rule
EXPONENTIAL & LOGARITHM
e^xe^xExponential
e^(ax)ae^(ax)Chain rule
a^xa^x · ln(a)General exponential
ln(x)1/xNatural log
ln(ax)1/xLog rule
logₐ(x)1/(x·ln(a))Log base a
TRIGONOMETRIC
sin(x)cos(x)Trig
cos(x)-sin(x)Trig
tan(x)sec²(x)Trig
sec(x)sec(x)·tan(x)Trig
cosec(x)-cosec(x)·cot(x)Trig
cot(x)-cosec²(x)Trig
INVERSE TRIGONOMETRIC
arcsin(x)1/√(1-x²)Inverse trig
arccos(x)-1/√(1-x²)Inverse trig
arctan(x)1/(1+x²)Inverse trig
COMBINATION RULES
f(x) + g(x)f'(x) + g'(x)Sum rule
f(x) - g(x)f'(x) - g'(x)Difference rule
c · f(x)c · f'(x)Constant multiple
f(x) · g(x)f'(x)g(x) + f(x)g'(x)Product rule
f(x) / g(x)[f'g − fg'] / g²Quotient rule
f(g(x))f'(g(x)) · g'(x)Chain rule

What is Differentiation?

Differentiation is one of the two fundamental operations of calculus (the other being integration). At its core, differentiation answers the question: how fast is this quantity changing? More precisely, it finds the instantaneous rate of change of a function at any given point — which is the same as the gradient (slope) of the curve at that point.

If you have a curve y = f(x), the derivative f'(x) — also written dy/dx — gives you the gradient of the tangent line to that curve at every value of x. When the curve is rising steeply, dy/dx is large and positive. When it is flat (at a maximum or minimum), dy/dx equals zero. When it slopes downward, dy/dx is negative.

Differentiation was developed independently by Isaac Newton and Gottfried Wilhelm Leibniz in the 17th century. Newton used dot notation (ẋ) mainly for physics, while Leibniz introduced the dy/dx notation used in mathematics and engineering today.

dy/dx Notation Explained

The notation dy/dx (read "dee y by dee x") means the derivative of y with respect to x. The "d" represents an infinitesimally small change. So dy/dx is the ratio of an infinitesimally small change in y to an infinitesimally small change in x — which is precisely the gradient of the curve.

Alternative notations for the derivative of f(x) include:

  • f'(x) — Lagrange or prime notation (most common in pure maths)
  • dy/dx — Leibniz notation (most common in applied maths and physics)
  • Df(x) — Euler's operator notation
  • — Newton's dot notation (mainly used in mechanics for time derivatives)

In A-Level mathematics, you will mostly encounter f'(x) and dy/dx. For the second derivative (the rate of change of the rate of change), you write f''(x) or d²y/dx².

The Power Rule: Differentiating x^n

The power rule is the most frequently used rule in differentiation and applies to any power of x:

Power Rule: If f(x) = x^n, then f'(x) = n · x^(n-1)
With constant: If f(x) = ax^n, then f'(x) = an · x^(n-1)

The steps are always: (1) bring the power down as a multiplier, (2) reduce the power by one. Examples:

  • d/dx(x³) = 3x² — power 3 comes down, new power is 2
  • d/dx(4x&sup5;) = 20x&sup4; — 4 times 5 = 20, new power is 4
  • d/dx(x) = 1 — because x = x¹, so 1·x&sup0; = 1
  • d/dx(7) = 0 — any constant differentiates to zero
  • d/dx(x^(-2)) = -2x^(-3) = -2/x³ — works for negative powers too
  • d/dx(√x) = d/dx(x^(1/2)) = (1/2)x^(-1/2) = 1/(2√x)

Worked Example: Differentiate f(x) = x³ + 2x² − 5x + 3

Apply the power rule to each term separately (sum rule):

d/dx(x³) = 3x²

d/dx(2x²) = 4x

d/dx(-5x) = -5

d/dx(3) = 0

Therefore: f'(x) = 3x² + 4x − 5

To find the gradient at x = 2: f'(2) = 3(4) + 4(2) − 5 = 12 + 8 − 5 = 15

The Chain Rule

The chain rule is used to differentiate composite functions — functions of functions. If y = f(g(x)), where g(x) is the "inner" function and f is the "outer" function:

Chain Rule: d/dx[f(g(x))] = f'(g(x)) · g'(x)

A practical way to remember it: differentiate the outer function (leaving the inner unchanged), then multiply by the derivative of the inner function.

Worked Example: Differentiate y = sin(3x²)

Outer function: sin(•) → derivative cos(•). Inner function: 3x² → derivative 6x.

dy/dx = cos(3x²) × 6x = 6x·cos(3x²)

Worked Example: Differentiate y = (2x + 1)&sup5;

Outer: (•)&sup5; → 5(•)&sup4;. Inner: (2x + 1) → 2.

dy/dx = 5(2x + 1)&sup4; × 2 = 10(2x + 1)&sup4;

The Product Rule

When two functions are multiplied together, use the product rule. If y = u(x) · v(x):

Product Rule: dy/dx = u'v + uv'
Memory aid: "first times derivative of second, plus second times derivative of first"

Worked Example: Differentiate y = x² · sin(x)

Let u = x² (so u' = 2x) and v = sin(x) (so v' = cos(x)).

dy/dx = (2x)(sin x) + (x²)(cos x) = 2x·sin(x) + x²·cos(x)

The Quotient Rule

When one function is divided by another, use the quotient rule. If y = f(x) / g(x):

Quotient Rule: dy/dx = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]²
Memory aid: "low d-high minus high d-low, square the bottom and away we go"

Worked Example: Differentiate y = sin(x) / x

f(x) = sin(x), f'(x) = cos(x). g(x) = x, g'(x) = 1.

dy/dx = [cos(x)·x − sin(x)·1] / x² = [x·cos(x) − sin(x)] / x²

Partial Differentiation Explained

When a function depends on more than one variable — for example f(x, y) or f(x, y, z) — we can differentiate with respect to each variable separately. This is called partial differentiation.

The notation changes from d to the curly ∂ symbol: ∂f/∂x means "the partial derivative of f with respect to x" — treating all other variables as constants.

Partial Differentiation Rule: When finding ∂f/∂x, treat every variable other than x as a constant and differentiate with respect to x using the normal rules.

Worked Example: Partial derivatives of f(x,y) = x²y + 3xy² + 2y

∂f/∂x (treat y as constant):

∂f/∂x = 2xy + 3y²

∂f/∂y (treat x as constant):

∂f/∂y = x² + 6xy + 2

Note: The term 2y differentiates to 0 when finding ∂f/∂x (because y is constant), but differentiates to 2 when finding ∂f/∂y.

Partial derivatives are essential in multivariable calculus, thermodynamics, economics (marginal analysis), and machine learning (gradient descent uses partial derivatives to minimise loss functions).

Differentiation in A-Level Maths

In A-Level Mathematics (both AQA and Edexcel), differentiation typically appears in:

  • C1/AS Pure: Power rule, differentiating polynomials, finding gradients and tangents
  • C2/A2 Pure: Chain rule, product rule, quotient rule, differentiating trig and exponentials
  • A2 Pure: Implicit differentiation, parametric differentiation, second derivatives
  • Mechanics: Velocity (dx/dt) and acceleration (d²x/dt²) as derivatives of position

Differentiation also appears in the applications of calculus: finding stationary points (where f'(x) = 0), determining whether they are maxima or minima (using f''(x) — positive means minimum, negative means maximum), and curve sketching.

Applications of Differentiation

Finding maximum and minimum points: Set f'(x) = 0 and solve for x. These are the stationary points. Check f''(x): if f''(x) < 0 it is a maximum, if f''(x) > 0 it is a minimum.

Velocity and acceleration: If s(t) is displacement as a function of time, then velocity v = ds/dt and acceleration a = dv/dt = d²s/dt².

Optimisation in business: If a company's profit P depends on output q, then dP/dq = 0 gives the profit-maximising output. This is marginal profit = 0.

Rates of change in science: In chemistry, reaction rates are derivatives of concentration with respect to time. In physics, the rate of heat flow is a partial derivative of temperature with respect to position.

Machine learning: Neural networks are trained using gradient descent, which repeatedly computes partial derivatives (the gradient) of the loss function with respect to all weights and biases.

Frequently Asked Questions

What is dy/dx? +

dy/dx is Leibniz notation for the derivative of y with respect to x. It represents the instantaneous rate of change of y as x changes — equivalently, the gradient of the curve y = f(x) at any given point. If y = x², then dy/dx = 2x, meaning at x = 3 the gradient of the parabola is 6.

How do you differentiate x^n? +

Use the power rule: multiply by the power, then reduce the power by one. d/dx(x^n) = n·x^(n-1). For example: d/dx(x&sup5;) = 5x&sup4;, d/dx(x²) = 2x, d/dx(x) = 1, d/dx(7) = 0. This works for any real power, including negative and fractional powers.

What is partial differentiation? +

Partial differentiation finds the derivative of a multivariable function with respect to one variable, treating all other variables as constants. The notation uses ∂ instead of d. For f(x,y) = x²y + 3y: ∂f/∂x = 2xy (y is constant, 3y drops out) and ∂f/∂y = x² + 3 (x² is the coefficient of y).

How do you use the chain rule? +

The chain rule applies to composite functions f(g(x)): d/dx[f(g(x))] = f'(g(x)) × g'(x). Step 1: differentiate the outer function (leaving the inner function unchanged). Step 2: multiply by the derivative of the inner function. Example: d/dx[e^(x²)] = e^(x²) × 2x = 2x·e^(x²).

What is the derivation of differentiation from first principles? +

From first principles, the derivative is defined as: f'(x) = lim(h→0) [f(x+h) − f(x)] / h. For f(x) = x²: [(x+h)² − x²] / h = [x² + 2xh + h² − x²] / h = [2xh + h²] / h = 2x + h. As h→0, this gives 2x, confirming the power rule.

What is the derivative of x/(1+x)? +

Use the quotient rule with f = x (f' = 1) and g = 1+x (g' = 1). d/dx[x/(1+x)] = [(1)(1+x) − (x)(1)] / (1+x)² = [1+x − x] / (1+x)² = 1/(1+x)². This result is useful in logistic functions and economics.

Second Derivatives and Higher-Order Differentiation

The second derivative, written f''(x) or d²y/dx², is the derivative of the derivative. It measures how fast the gradient itself is changing — in other words, the rate of change of the rate of change. The second derivative is used to:

  • Classify stationary points: If f'(a) = 0 and f''(a) < 0, the point is a local maximum. If f''(a) > 0, it is a local minimum. If f''(a) = 0, the second derivative test is inconclusive.
  • Determine concavity: f''(x) > 0 means the curve is concave up (bowl-shaped). f''(x) < 0 means concave down (arch-shaped).
  • Points of inflection: Where f''(x) changes sign, there is a point of inflection — where the concavity switches.

Worked Example: Find and classify stationary points of y = x³ − 6x² + 9x + 1

Step 1: Find dy/dx

dy/dx = 3x² − 12x + 9

Step 2: Set dy/dx = 0

3x² − 12x + 9 = 0 ⇒ x² − 4x + 3 = 0 ⇒ (x−1)(x−3) = 0
x = 1 or x = 3

Step 3: Find d²y/dx²

d²y/dx² = 6x − 12

Step 4: Classify

At x = 1: d²y/dx² = 6(1) − 12 = −6 < 0 ⇒ LOCAL MAXIMUM
At x = 3: d²y/dx² = 6(3) − 12 = +6 > 0 ⇒ LOCAL MINIMUM

y-values: y(1) = 1 − 6 + 9 + 1 = 5 (maximum). y(3) = 27 − 54 + 27 + 1 = 1 (minimum).

Implicit Differentiation

Sometimes a curve is defined implicitly, meaning y is not expressed explicitly as a function of x. For example, the circle x² + y² = 25 defines y implicitly. To differentiate implicitly, differentiate both sides with respect to x, treating y as a function of x and using the chain rule whenever y appears.

Worked Example: Find dy/dx for x² + y² = 25

Differentiate both sides with respect to x:

d/dx(x²) + d/dx(y²) = d/dx(25)
2x + 2y · (dy/dx) = 0
dy/dx = −x/y

So at the point (3, 4) on the circle: dy/dx = −3/4. The gradient of the tangent is −3/4.

Parametric Differentiation

When a curve is defined parametrically — both x and y as functions of a parameter t — the derivative dy/dx is found using:

Parametric Differentiation: dy/dx = (dy/dt) ÷ (dx/dt)

Worked Example: Find dy/dx for x = t², y = 2t

dx/dt = 2t
dy/dt = 2
dy/dx = (dy/dt) ÷ (dx/dt) = 2 ÷ (2t) = 1/t

This is the parametric form of a parabola y² = 4x, and the gradient at parameter t is 1/t.

Logarithmic Differentiation

Logarithmic differentiation is a technique used when the function is too complex for direct rules — especially useful for functions of the form y = [f(x)]^[g(x)], where both the base and exponent contain x.

Method: Take the natural log of both sides, simplify using log laws, differentiate implicitly, then multiply both sides by y.

Worked Example: Differentiate y = x^x

ln(y) = x·ln(x)
(1/y)·(dy/dx) = ln(x) + x·(1/x) = ln(x) + 1
dy/dx = y·[ln(x) + 1] = x^x·[ln(x) + 1]

Common Mistakes and Tips

Common Mistake Correct Approach Example
Forgetting to use chain rule on composite functions Identify inner function; multiply by its derivative d/dx[sin(2x)] = 2cos(2x), NOT cos(2x)
Differentiating a constant to 1 instead of 0 Constants always differentiate to 0 d/dx[7] = 0, NOT 1
Confusing product rule and sum rule Sum rule applies to f+g; product rule to f×g d/dx[x²·sin(x)] requires product rule
Dropping the exponent after power rule New power = old power − 1 d/dx[x&sup5;] = 5x&sup4;, NOT 5x&sup5;
Wrong sign in quotient rule f'g − fg' (numerator order matters) d/dx[x/sin(x)] = [sin(x) − x·cos(x)] / sin²(x)
Confusing d/dx[e^x] = e^x with d/dx[a^x] d/dx[a^x] = a^x · ln(a) d/dx[2^x] = 2^x · ln(2)
Treating partial ∂f/∂x as full derivative Keep other variables constant in partial differentiation ∂/∂x[xy²] = y² (y is constant)