Calculus

Calculus Solver with Step-by-Step Answers

Derivatives, integrals, limits, and series worked line by line, with the rule that justifies each line named as it is used.

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Calculus questions are rarely lost on the calculus. They are lost on the algebra underneath it: a sign dropped when a bracket is subtracted, a factor of $3$ left behind because the inner derivative was forgotten, an antiderivative that is right except for the constant. The working here is deliberately granular for that reason — the differentiation of a product is split from the chain rule applied inside it, so you can see which half broke.

Type the function however is convenient: x^2 sin(3x), integral of x*e^(2x), lim (sin 5x)/(2x) as x->0. If your course requires a specific technique, say so, and the solution takes that route even when another is shorter.

What this solver handles

Derivatives by power, product, quotient, and chain rule, including nested compositions three layers deep.
Implicit differentiation, logarithmic differentiation, and related rates with the constraint equation stated first.
Indefinite and definite integrals by substitution, with limits converted rather than substituted back.
Integration by parts, including repeated applications and the circular case that has to be solved algebraically.
Limits at a point and at infinity, one-sided limits, and the indeterminate forms that need algebra before L'Hopital applies.
Curve sketching: critical points, the first and second derivative tests, inflection points, and asymptotes.
Optimisation and applied maxima, with the domain of the physical variable made explicit.
Areas between curves, volumes by discs, washers, and shells, and arc length.
Sequences and series: convergence tests, power series, Taylor and Maclaurin expansions.

Bracket generously when you type. sin x^2 could mean the sine of $x^{2}$ or the square of $sin x$ , and the two have entirely different derivatives. Give the limits with the variable on a definite integral, say which index a series starts at, and name the variable you are differentiating with respect to when more than one letter appears. Ambiguity in the question is the one error the working cannot catch, because the solution will be perfectly correct for the problem it was handed.

Three problems, worked

One from each half of a first course, plus the limit that motivates the second derivative of the cosine. Each is short enough to follow without a pencil and long enough to show where the rule is actually invoked.

Worked example

\frac{d}{d x} [x^{2} sin (3 x)]

1
This is a product of two functions of x, so the product rule applies. Label the factors before differentiating anything.
$u = x^{2}, v = sin (3 x)$
2
The derivative of u is straightforward. The derivative of v needs the chain rule: differentiate the sine, then multiply by the derivative of the inside.
$u^{'} = 2 x, v^{'} = 3 cos (3 x)$
3
Assemble with the product rule: derivative of the first times the second, plus the first times the derivative of the second.
$u^{'} v + u v^{'} = 2 x sin (3 x) + x^{2} \cdot 3 cos (3 x)$
4
Tidy the second term. The factor of 3 is the inner derivative — losing it is the single most common error on this type.
$2 x sin (3 x) + 3 x^{2} cos (3 x)$

Answer

2 x sin (3 x) + 3 x^{2} cos (3 x)

Worked example

\int x e^{2 x} d x

1
A polynomial multiplied by an exponential: parts, with the polynomial as u because differentiating it makes it simpler.
$u = x, d v = e^{2 x} d x$
2
Differentiate u and integrate dv. The one-half comes from the chain rule running backwards on the exponential.
$d u = d x, v = \frac{1}{2} e^{2 x}$
3
Apply the parts formula: uv minus the integral of v du.
$\int x e^{2 x} d x = \frac{1}{2} x e^{2 x} - \int \frac{1}{2} e^{2 x} d x$
4
The remaining integral has no x attached, which is the point of the choice. Integrate it directly.
$\int \frac{1}{2} e^{2 x} d x = \frac{1}{4} e^{2 x}$
5
Subtract, and add the constant. Differentiating the result gives one-half e^(2x) + x e^(2x) − one-half e^(2x), which collapses back to the integrand.
$\frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C$

Answer

\frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C

Worked example

lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}

1
Substitute first, always. Both numerator and denominator go to zero, so the expression is indeterminate and the limit is not yet decided.
$\frac{1 - cos 0}{0 ^{2}} = \frac{0}{0}$
2
Zero over zero permits L'Hopital's rule. Differentiate the top and the bottom separately — this is not the quotient rule.
$x \to 0 lim \frac{sin x}{2 x}$
3
Substituting again still gives zero over zero, so the rule applies a second time.
$x \to 0 lim \frac{cos x}{2}$
4
Now substitution works: cosine of zero is 1.
$\frac{1}{2}$
5
Sanity check with the Maclaurin series. Near zero, cos x is approximately 1 − x²/2, so the numerator is about x²/2 and the ratio is about one-half — independent of x, which is exactly why the limit exists.
$\frac{1 - ( 1 - \frac{x ^{2}}{2} )}{x ^{2}} = \frac{1}{2}$

Answer

\frac{1}{2}

Calculus answers are unusually cheap to verify, which makes skipping the verification an expensive habit. An antiderivative can be differentiated: doing that to $\frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x}$ gives $\frac{1}{2} e^{2 x} + x e^{2 x} - \frac{1}{2} e^{2 x}$ , and the two halves cancel to leave the original integrand. That confirms the result no matter how the parts were chosen.

A limit can be sampled instead. Putting $x = 0.1$ into the third problem gives 0.4996, near enough to one half to settle any doubt about applying the rule twice. A derivative can be checked the same way, against a difference quotient with a small step. None of these is a proof, but each takes under a minute and catches the arithmetic slips that cost most of the marks lost on a calculus paper.

Choosing an integration technique

Differentiation is an algorithm; integration is a search. The difference is why students can differentiate anything after a fortnight and still stall on integrals in month three. Narrow the search by looking at the shape of the integrand before touching it, and back out of a branch quickly when the new integral is worse than the old one.

Which technique should you try first?

What does the integrand look like?

If two branches look plausible, spend thirty seconds on the cheaper one. Substitution costs one line to test: write down $u$ , compute $d u$ , and see whether the rest of the integrand is a constant multiple of it. Parts costs a full application before you know.

Where students go wrong

The missing inner derivative

$\frac{d}{d x} sin (3 x)$ is $3 cos (3 x)$ . The factor of 3 is the derivative of the inside, and dropping it is the most common single error in first-year calculus. A quick guard: if the inside is anything other than plain $x$ , there must be an extra factor in your answer.

Differentiating a product factor by factor

The derivative of $x^{2} sin x$ is not $2 x cos x$ . Derivatives do not distribute over multiplication. If you ever doubt it, test with $f (x) = x \cdot x$ : the true derivative is $2 x$ , while multiplying the derivatives gives 1.

Applying L'Hopital to a form that is not indeterminate

The rule needs zero over zero or infinity over infinity. For $lim_{x \to 0} \frac{x + 1}{x ^{2} + 1}$ plain substitution gives 1, but differentiating top and bottom gives $\frac{1}{2 x}$ , which diverges. Always substitute before reaching for the rule.

Substituting without moving the limits

After changing variable in a definite integral, the numbers on the integral sign refer to $x$ , not $u$ . Either convert them using the substitution or convert the antiderivative back before evaluating. Doing neither produces a plausible-looking number that is wrong, which is worse than an obvious error. The same care applies to one-sided limits, where the direction of approach changes with the substitution.

Find the slip in this quotient rule

One of these lines is wrong. Click it.

Three of those four are algebra errors wearing calculus notation, which is the usual pattern at this level: the rule is known and the execution slips. The exception is L'Hopital, where a valid rule is applied in a situation that does not license it. If integrals specifically keep stalling, the trouble is method selection rather than technique, and the guide to choosing an integration method walks down each branch including the ones worth abandoning early.

Formulas worth knowing cold

Calculus reference

Tap any formula with a derivation to see where it comes from.

Differentiation

Integration

Limits and series

Practice

Four to try

Answers are checked here — nothing is sent anywhere.

1
$\frac{d}{d x} (2 x + 1)^{5}$
2
$\int 6 x^{2} d x$
3
$lim_{x \to 0} \frac{s i n 4 x}{x}$
4
$f (x) = x^{3} - 4 x, find f^{'} (1)$

Frequently asked questions

Is the calculus solver free?

The first solution is free with no account. A free account then gives three solutions a day, which covers a normal problem set spread over a week. Gauth Plus is $11.99 a month with a three-day free trial and removes the daily limit.

Does it show the steps or just the answer?

Each line names the rule it used — chain rule on the inner function, parts with u equal to the polynomial, L'Hopital because the form was zero over zero. An antiderivative with no working attached is worth almost nothing on a marked script.

Can it do definite integrals and change the limits?

Yes. On a substitution it converts the limits to the new variable rather than substituting back, and both routes are shown when they differ in difficulty.

Will it handle implicit differentiation and related rates?

Yes. Implicit problems keep the dy/dx factors visible at every step, and related-rates solutions state the relationship between the variables before differentiating with respect to time.

Can I ask why a particular technique was chosen?

Ask a follow-up on the step. Answering why parts beat substitution on a given integrand is usually more valuable than the antiderivative itself, and follow-ups do not use up your quota.

Does it cover sequences and series?

Convergence tests, radius of interval of convergence, and Taylor or Maclaurin expansions are covered, with the reason a particular test was chosen stated before it is applied.

Keep going

The chain rule, explained plainlySpotting a composite function and keeping track of the layers.Integration by parts and LIATEChoosing u correctly, and recognising the circular case.Limits and continuity: intuition firstWhat a limit means, and the three conditions for continuity.Precalculus solverFor the algebra and trig that calculus assumes you already own.Derivative calculator with stepsDifferentiate anything and see which rule fires at each stage.