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Free Derivative Calculator with Steps

Calculate derivatives step-by-step. Learn power rule, chain rule, product rule, and quotient rule with instant feedback. Analyse critical points, tangent lines, and function behaviour.

Step-by-step learning with explanations

Variable:
Order:

Enter an Expression

Enter a mathematical expression above to learn how to differentiate it step-by-step.

Reference

Examples

(constant)
(x = x¹)
(general)

Special Cases

What is a Derivative?

A derivative measures how a function changes as its input changes. Written as f'(x) or dy/dx, it represents the instantaneous rate of change or the slope of the tangent line at any point.

Formally, the derivative is defined from first principles as:

f'(x) = lim(h→0) [f(x+h) - f(x)] / h

Geometric interpretation: The derivative at a point gives the slope of the tangent line to the curve at that point. A positive derivative means the function is increasing; a negative derivative means it's decreasing; a zero derivative means the tangent is horizontal (stationary point).

Real-world applications: Derivatives model velocity (rate of change of position), acceleration (rate of change of velocity), marginal cost in economics, population growth rates in biology, and optimisation problems across all sciences and engineering.

Differentiation Rules & Methods

Power Rule

d/dx[xⁿ] = nxⁿ⁻¹

Bring the exponent down, reduce by 1. Works for all real exponents.

e.g. d/dx[x³] = 3x²

Chain Rule

d/dx[f(g(x))] = f'(g(x))·g'(x)

For composite functions: derivative of outer times derivative of inner.

e.g. d/dx[sin(x²)] = cos(x²)·2x

Product Rule

d/dx[fg] = f'g + fg'

"First d-second + second d-first" for multiplied functions.

e.g. d/dx[x·sin(x)] = sin(x) + x·cos(x)

Quotient Rule

d/dx[f/g] = (f'g - fg')/g²

"Low d-high minus high d-low, over the square of what's below."

e.g. d/dx[x/(x+1)] = 1/(x+1)²

Trigonometric

sin→cos, cos→-sin, tan→sec²

Standard derivatives for all 6 trig functions.

e.g. d/dx[cos(x)] = -sin(x)

Exponential & Logarithmic

eˣ→eˣ, ln(x)→1/x

eˣ is its own derivative. For aˣ, multiply by ln(a).

e.g. d/dx[e²ˣ] = 2e²ˣ

Critical Points & Curve Sketching

Critical points are where f'(x) = 0 or f'(x) is undefined. These are the potential maxima, minima, or inflection points of a function. Finding them is essential for curve sketching and optimisation problems.

The second derivative test classifies each critical point:

f''(x)ClassificationShape
> 0Local MinimumConcave up (U-shape)
< 0Local MaximumConcave down (n-shape)
= 0InconclusiveNeed further analysis

Inflection points are where the concavity changes: f''(x) = 0 and f''(x) changes sign. At an inflection point, the curve transitions from concave up to concave down (or vice versa).

Curve sketching process:

  1. Find f'(x) and solve f'(x) = 0 for critical points
  2. Use the second derivative test to classify each point
  3. Determine increasing/decreasing intervals from the sign of f'(x)
  4. Find inflection points where f''(x) changes sign
  5. Note y-intercept f(0) and any x-intercepts
  6. Sketch the curve using all this information

Use our Analyse tab to compute all of this automatically!

Common Derivatives Reference Table

Basic Functions

d/dx[c]= 0
d/dx[x]= 1
d/dx[xⁿ]= nxⁿ⁻¹
d/dx[√x]= 1/(2√x)
d/dx[1/x]= -1/x²

Trigonometric

d/dx[sin(x)]= cos(x)
d/dx[cos(x)]= -sin(x)
d/dx[tan(x)]= sec²(x)
d/dx[sec(x)]= sec(x)tan(x)
d/dx[csc(x)]= -csc(x)cot(x)
d/dx[cot(x)]= -csc²(x)

Exp, Log & Inverse Trig

d/dx[]=
d/dx[]= aˣ·ln(a)
d/dx[ln(x)]= 1/x
d/dx[arcsin(x)]= 1/√(1-x²)
d/dx[arccos(x)]= -1/√(1-x²)
d/dx[arctan(x)]= 1/(1+x²)

With the chain rule: d/dx[sin(kx)] = k·cos(kx), d/dx[e^(kx)] = k·e^(kx), d/dx[ln(kx)] = 1/x

Common Differentiation Mistakes

Forgetting the chain rule inner derivative

d/dx[sin(3x)] = cos(3x)

d/dx[sin(3x)] = 3cos(3x)

Always multiply by the derivative of the inner function.

Wrong sign on cos derivative

d/dx[cos(x)] = sin(x)

d/dx[cos(x)] = -sin(x)

Remember: cosine goes negative. The minus sign is essential.

Multiplying derivatives for product rule

d/dx[x·sin(x)] = 1·cos(x)

d/dx[x·sin(x)] = sin(x) + x·cos(x)

Use f'g + fg', not f'·g'.

Forgetting to reduce the exponent

d/dx[x³] = 3x³

d/dx[x³] = 3x²

Power rule: bring down n, then reduce to n-1.

Confusing d/dx[e^x] with d/dx[x^e]

d/dx[x^e] = x^e (treating as exponential)

d/dx[x^e] = e·x^(e-1) (power rule)

Variable in exponent = exponential. Variable in base = power rule.

Not simplifying after quotient rule

Leaving (f'g - fg')/g² unsimplified

Always expand, cancel common factors, and simplify

Examiners expect a simplified final answer.

Worked Examples

GCSEPower Rule

f(x) = 3x² - 4x + 7

f(x) = 3x² - 4x + 7

Differentiate term by term:

d/dx[3x²] = 3 × 2 × x²⁻¹ = 6x

d/dx[-4x] = -4 × 1 × x¹⁻¹ = -4

d/dx[7] = 0

f'(x) = 6x - 4

A-LevelChain Rule

f(x) = sin(3x)

f(x) = sin(3x)

Outer: sin(u), Inner: u = 3x

Outer derivative: cos(u)

Inner derivative: 3

Chain rule: cos(3x) × 3

f'(x) = 3cos(3x)

A-LevelProduct Rule

f(x) = x²·eˣ

f(x) = x² × eˣ

Let f = x², g = eˣ

f' = 2x, g' = eˣ

Product rule: f'g + fg'

= 2x·eˣ + x²·eˣ

f'(x) = eˣ(2x + x²) = x·eˣ(2 + x)

A-Level FurtherQuotient Rule

f(x) = (x² + 1)/(x - 1)

f(x) = (x² + 1)/(x - 1)

f = x² + 1, g = x - 1

f' = 2x, g' = 1

Quotient rule: (f'g - fg')/g²

= (2x(x-1) - (x²+1)(1))/(x-1)²

= (2x² - 2x - x² - 1)/(x-1)²

f'(x) = (x² - 2x - 1)/(x - 1)²

Exam Tips for Differentiation

1

State the rule you're using

Write "Using the chain rule:" or "Product rule:" before applying. This earns method marks even if you make an arithmetic slip.

2

Show the "before and after" for chain rule

Clearly write the outer derivative and inner derivative separately before combining. Examiners look for this structure.

3

Always simplify your final answer

Factor out common terms, simplify fractions, and write in the neatest form. Unsimplified answers may lose marks.

4

Check sign on trig derivatives

sin→+cos, cos→-sin. The negative on cos→-sin is the #1 source of sign errors in differentiation.

5

For optimisation: set f'(x) = 0

When asked to find maximum/minimum values, differentiate, set equal to zero, solve for x, then substitute back to find y.

6

Verify with a simple substitution

Plug in x = 0 or x = 1 into both the original and derivative to sanity-check your answer. This catches many errors.

Frequently Asked Questions

How do I find the derivative of a function?

Identify the function type and apply the appropriate rule: power rule for polynomials, chain rule for composite functions, product/quotient rules for multiplied/divided functions. Our calculator shows each step of the process.

What is the chain rule and when do I use it?

The chain rule differentiates composite functions: d/dx[f(g(x))] = f'(g(x)) · g'(x). Use it whenever one function is inside another, like sin(x²) or e^(3x).

What is the derivative of sin(x)?

The derivative of sin(x) is cos(x). For sin(kx), apply the chain rule to get k·cos(kx). Similarly, d/dx[cos(x)] = -sin(x).

What is the derivative of e^x?

The derivative of eˣ is eˣ itself — the only function that equals its own derivative! For e^(kx), use chain rule: k·e^(kx).

What is the derivative of ln(x)?

The derivative of ln(x) is 1/x. For ln(f(x)), apply the chain rule: f'(x)/f(x).

What is the product rule?

The product rule differentiates products: d/dx[f·g] = f'·g + f·g'. For example, d/dx[x·sin(x)] = sin(x) + x·cos(x).

What is the quotient rule?

The quotient rule: d/dx[f/g] = (f'g - fg')/g². Mnemonic: "low d-high minus high d-low, over the square of what's below."

How do I find critical points of a function?

Set f'(x) = 0 and solve. Classify using the second derivative test: f''(x) > 0 = min, f''(x) < 0 = max. Use our Analyse tab for automatic computation.

What is the second derivative test?

At a critical point where f'(x₀) = 0: if f''(x₀) > 0, it's a local minimum (concave up); if f''(x₀) < 0, it's a local maximum (concave down); if f''(x₀) = 0, the test is inconclusive.

Can this calculator find higher-order derivatives?

Yes! Select the derivative order (2nd, 3rd, etc.) and the calculator will differentiate repeatedly, showing each step. Used for Taylor series and concavity analysis.

What is implicit differentiation?

Implicit differentiation is used when y isn't explicitly solved for x. Differentiate both sides with respect to x, applying the chain rule to y terms, then solve for dy/dx.

Is this suitable for A-Level / GCSE maths?

Absolutely! Covers all A-Level differentiation: power, chain, product, quotient rules, trig/exponential derivatives, and critical point analysis. The Learn Mode teaches exam-ready methods.

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