Symbolic Derivatives and Integrals With Steps
Calculate derivatives and integrals symbolically with step-by-step solutions and interactive graphs
Calculate derivatives and integrals symbolically with step-by-step solutions and interactive graphs
The page handles three operations: the derivative d/dx[f(x)], the indefinite integral ∫f(x)dx + C, and the definite integral ∫ₐᵇ f(x)dx. Symbolic, not numeric. The engine returns expressions, not floating-point values, except when an integrand has no elementary antiderivative and a definite-bounds query falls back to adaptive quadrature.
Coverage on the differentiation side is complete for elementary functions: power, product, quotient, and chain rules across trig, inverse trig, exponential, and logarithmic forms. Integration handles polynomials, basic trig, exponentials, and the standard substitution / by-parts patterns; ∫e^(−x²)dx comes back as erf(·) because it has no elementary closed form, which is the correct answer rather than a bug. Two things that catch first-time users: indefinite integrals always carry +C (drop it and you lose the boundary-condition information that matters in ODE work), and parser syntax wants explicit multiplication (write 3*x, not 3x) plus parentheses around fractions ((x+1)/(x−2), not x+1/x−2, which gets read by operator precedence and not the way you meant).
Most input errors trace back to syntax. Write multiplication explicitly: use 3*x, not 3x. Exponents use the caret: x^2 for x squared, e^x for the exponential. Parentheses matter—sin(2*x) differs from sin2*x (which the parser may reject or misread).
Common functions: sin, cos, tan for trig; ln for natural log; sqrt for square root; exp for e^(...). Inverse trig uses asin, acos, atan. Constants are pi and e. Some parsers accept log as natural log; others treat log as base-10. Check which convention applies.
If your expression involves fractions, wrap numerator and denominator in parentheses: (x^2 + 1)/(x - 3). Otherwise the parser may read x^2 + 1/x - 3, which is very different. Test by entering simple known functions first (like x^2) to confirm parsing works as expected.
Examples:
• x^3 - 4*x + 7
• sin(x)*cos(x)
• e^(-x^2)
• ln(x + 1)/(x^2)
The power rule handles polynomials: d/dx[xⁿ] = n·xⁿ⁻¹. For x³, the derivative is 3x². Constants vanish: d/dx[7] = 0. When functions multiply, the product rule kicks in: d/dx[u·v] = u′v + uv′.
Composite functions require the chain rule: d/dx[f(g(x))] = f′(g(x))·g′(x). For sin(x²), outer derivative cos(...) multiplies inner derivative 2x, yielding 2x·cos(x²). Most errors arise from forgetting that inner derivative.
The quotient rule applies when dividing: d/dx[u/v] = (u′v − uv′)/v². Alternatively, rewrite as u·v⁻¹ and use product plus chain rules. The tool simplifies automatically, but raw outputs may look different from textbook forms—both can be correct.
Verification tip: Differentiate your integral to recover the original function. If f(x) integrates to F(x), then F′(x) should equal f(x). This sanity check catches most errors.
Indefinite integrals produce a family of functions differing by a constant. ∫2x dx = x² + C. That + C matters—it represents all vertical shifts of the antiderivative. Omitting C loses information about initial conditions or boundary values.
Definite integrals evaluate between bounds: ∫₀³ 2x dx = [x²]₀³ = 9 − 0 = 9. The result is a number, not a function. It measures net signed area under the curve from x = 0 to x = 3. Negative regions subtract from positive regions.
Some functions lack elementary antiderivatives. ∫e^(−x²)dx has no closed form using standard functions—it's expressed via the error function erf. If the tool returns a special function or a numerical approximation, that's not an error; it's the nature of the integral.
Fundamental Theorem of Calculus:
∫ₐᵇ f(x) dx = F(b) − F(a), where F′(x) = f(x)
Plotting catches mistakes that symbolic manipulation hides. If f(x) has a local maximum at x = 2, the derivative f′(x) should cross zero there—going from positive to negative. If your computed derivative doesn't show this, something's wrong.
For integrals, the antiderivative F(x) should increase where f(x) is positive and decrease where f(x) is negative. Inflection points on F correspond to extrema on f. Overlay f and F on the same plot to verify this behavior visually.
Definite integrals show as shaded area under the curve. Positive regions above the axis contribute positively; negative regions below subtract. If your numerical result seems off, check if sign cancellations explain the discrepancy.
Watch for: Asymptotes (vertical lines where f blows up), discontinuities, and domain restrictions. Integrating across a vertical asymptote requires improper integral handling—the tool may return infinity or refuse.
∫(1/x)dx = ln|x| + C, not ln(x). The absolute value matters because 1/x is defined for negative x, and ln of a negative number is undefined in reals. Some tools omit the absolute value bars; mentally add them for negative domains.
Trig integrals often require identities. ∫sin²(x)dx doesn't simplify directly—you need sin²x = (1 − cos(2x))/2 first. Similarly, ∫sec²(x)dx = tan(x) + C is a memorized result. If your output looks unfamiliar, check if a trig identity bridges the gap.
Different but equivalent forms abound. −cos(x) + C and sin(x − π/2) + C represent the same antiderivative of sin(x). If your answer differs from the textbook's, graph both—they may coincide up to a constant shift.
Identity reminder: sin²x + cos²x = 1; 1 + tan²x = sec²x; cos(2x) = cos²x − sin²x. These unlock many integrals that look impossible at first glance.
Why does my integral have + C but the definite integral doesn't?
Indefinite integrals give a family of antiderivatives, all differing by a constant. Definite integrals subtract F(a) from F(b), which cancels the constant. The result is a specific number, not a function.
The derivative looks different from my textbook answer. Is it wrong?
Not necessarily. Algebraic simplification can produce equivalent but different-looking forms. Graph both or substitute test values. If they match everywhere, both answers are correct.
What does it mean if the tool returns "erf" or "Si"?
These are special functions defined as integrals that lack elementary closed forms. erf is the error function; Si is the sine integral. They're perfectly valid answers—just not expressible using polynomials, trig, logs, and exponentials alone.
How do I handle improper integrals?
Improper integrals have infinite bounds or integrand singularities. Replace ∞ with a limit: ∫₁^∞ (1/x²)dx = lim(b→∞)[−1/x]₁ᵇ = 1. If the limit diverges, the integral doesn't exist. Tools may return "divergent" or infinity.
Can I differentiate piecewise functions?
Yes, but handle each piece separately and check continuity at boundaries. At transition points, the derivative may not exist (corner) or may have a jump discontinuity. Most simple tools don't support piecewise input directly.
Six problems with the steps written out. Standard textbook techniques: power rule, FTC, integration by parts, substitution, absolute values, and chain rule.
Problem. Compute d/dx ∫_a^x f(t) dt.
Fundamental Theorem of Calculus, Part 1. If f is continuous on [a, b] and F(x) = ∫_a^x f(t) dt, then F'(x) = f(x). The constant a doesn't matter (it shifts F by a constant, which differentiates to 0). The chain-rule generalization: d/dx ∫_a^(g(x)) f(t) dt = f(g(x)) · g'(x), used whenever the upper limit is itself a function of x.
Concrete example. d/dx ∫_0^(x²) sin(t) dt = sin(x²) · 2x by the chain rule, because g(x) = x² and g'(x) = 2x. If both limits depend on x, split the integral at any constant in between and apply the rule twice (once with a sign flip): d/dx ∫_(h(x))^(g(x)) f(t) dt = f(g(x)) · g'(x) − f(h(x)) · h'(x).
Answer: d/dx ∫_a^x f(t) dt = f(x).
Problem. Evaluate ∫_(-4)^(-1) 3·|x + 3| dx.
The absolute value flips at x = -3. Split the integral at the kink. On [-4, -3], x + 3 ≤ 0, so |x + 3| = -(x + 3). On [-3, -1], x + 3 ≥ 0, so |x + 3| = x + 3.
Left = ∫_(-4)^(-3) 3·-(x + 3) dx
= -3 · [x²/2 + 3x] from -4 to -3
= -3 · [(9/2 - 9) - (8 - 12)]
= -3 · [(-9/2) - (-4)]
= -3 · (-1/2)
= 1.5
Right = ∫_(-3)^(-1) 3·(x + 3) dx
= 3 · [x²/2 + 3x] from -3 to -1
= 3 · [(1/2 - 3) - (9/2 - 9)]
= 3 · [(-5/2) - (-9/2)]
= 3 · 2
= 6
Total = 1.5 + 6 = 7.5Answer: ∫_(-4)^(-1) 3·|x + 3| dx = 7.5.
Problem. Evaluate ∫ x · e^x dx.
Integration by parts: ∫ u dv = uv − ∫ v du. Pick u = x and dv = e^x dx, so du = dx and v = e^x.
∫ x · e^x dx = x · e^x − ∫ e^x dx
= x · e^x − e^x + C
= e^x (x − 1) + C
Check: d/dx [e^x (x − 1)] = e^x (x − 1) + e^x · 1 = x · e^x ✓Pick u in "LIATE" order (logs, inverse trig, algebraic, trig, exp) when in doubt. The polynomial here is the algebraic factor, which beats the exponential, so u = x is the right call.
If you accidentally pick u = e^x and dv = x dx, you get du = e^x dx and v = x²/2, leaving you with ∫ x²·e^x / 2 dx, which is harder than the original. That's the diagnostic: if integration by parts makes the remaining integral worse, swap your u and dv choices.
Answer: ∫ x · e^x dx = e^x (x − 1) + C.
Problem. Evaluate ∫ 2x · cos(x²) dx.
The factor 2x is exactly d/dx(x²), so substitute u = x², du = 2x dx. The integral becomes ∫ cos(u) du = sin(u) + C. Substitute back: sin(x²) + C.
The 2x being there "for free" is the giveaway that u-substitution is the right move. Without it, ∫ cos(x²) dx is the Fresnel integral, which has no elementary antiderivative.
Answer: ∫ 2x · cos(x²) dx = sin(x²) + C.
Problem. Evaluate ∫_0^1 x² dx.
Power rule: ∫ x^n dx = x^(n+1)/(n+1) + C for n ≠ -1. Here n = 2.
∫_0^1 x² dx = [x³/3] from 0 to 1
= 1/3 − 0
= 1/3 ≈ 0.3333The area under y = x² between 0 and 1 is exactly a third of the unit square. The generalization ∫_0^1 x^n dx = 1/(n+1) works for any n > -1. At n = -1 the integrand becomes 1/x and the integral diverges at 0, so the formula breaks at the boundary case (the antiderivative becomes ln|x| there instead).
Answer: ∫_0^1 x² dx = 1/3 ≈ 0.3333.
Problem. Compute d/dx [e^(x²)].
Chain rule. With outer f(u) = e^u and inner g(x) = x², f'(u) = e^u and g'(x) = 2x. So d/dx [e^(x²)] = e^(x²) · 2x = 2x · e^(x²).
Answer: d/dx [e^(x²)] = 2x · e^(x²). Differentiating sin(x²) (the u-sub answer above) gives 2x · cos(x²) by the same pattern, which is how derivative and integral check each other.
Not every function has an elementary antiderivative. ∫e^(−x²)dx comes back as erf because no combination of polynomials, exponentials, logs, and trig functions equals the integral. The page returns special functions or numerical approximations in those cases. That's the correct answer, not a bug.
Domain assumptions: ln(x) and √x evaluate on their natural real domains. Complex-valued extensions need a CAS that does branch-cut tracking properly.
Numerical fallback for definite integrals: highly oscillatory or singular integrands fall back to adaptive quadrature with finite precision. Improper integrals require explicit limit handling.
Parser quirks: the expression parser doesn't do implicit multiplication. Write 3*x, not 3x. Wrap fractions in parentheses to avoid operator-precedence surprises.
Note: For symbolic work at scale, Wolfram Alpha and Mathematica are the standard references. Maple is the equivalent in academic engineering settings. SymPy is the open-source workhorse and pairs cleanly with NumPy for numerical fallbacks. Treat outputs from this page as a homework check, not a paper-ready derivation.
Formulas and rules follow standard calculus references:
The page shows the symbolic answer plus the rule applied at each step. For u-substitution, you'll see the substitution chosen, the differential, and the back-substitution at the end. For integration by parts, the choice of u and dv is shown along with the resulting formula. ∫xe^x dx, for example, walks through u = x, dv = e^x dx, du = dx, v = e^x, then xe^x − ∫e^x dx = (x − 1)e^x + C. For more complex integrals, multiple steps are listed in order.
Two forms of FTC. Part 1: d/dx [∫ₐˣ f(t) dt] = f(x). The derivative of an integral with variable upper limit returns the integrand at that limit. Part 2 is the evaluation form: ∫ₐᵇ f(x) dx = F(b) − F(a) for any antiderivative F. If the upper bound is itself a function of x, apply the chain rule: d/dx [∫ₐ^g(x) f(t) dt] = f(g(x)) · g'(x). For ∫_{x²}^{x³} f(t) dt the derivative is f(x³)·3x² − f(x²)·2x.
Same object, different framing. An antiderivative of f is any F with F'(x) = f(x). The indefinite integral ∫f(x) dx denotes the family of all antiderivatives, written as F(x) + C with C arbitrary. Technically: every indefinite integral is an antiderivative plus a constant of integration, while "the antiderivative" usually drops the constant. In practice the terms are used interchangeably, and the +C matters because it encodes initial conditions in differential equations.
Both. For definite integrals ∫ₐᵇ f(x) dx, the page applies FTC Part 2: find an antiderivative F symbolically, then evaluate F(b) − F(a). For derivatives of integrals d/dx [∫ₐˣ f(t) dt], the page applies FTC Part 1 directly, returning f(x), with the chain rule on the upper bound if it's a function of x. When no elementary antiderivative exists (e.g., ∫e^(−x²) dx), the page falls back to numerical quadrature for the definite case and returns a special function (erf, Si, Ei) for the indefinite case.
Yes. Standard notation: sin, cos, tan, sec, csc, cot for trig; asin, acos, atan for inverses; ln (natural log) and log (also natural by default, with log10 for base 10). Use exp(x) or e^x for exponentials. The parser handles compositions like sin(2x), e^(−x²), ln(1 + x²) directly. Two syntax notes that catch people: write 3*x, not 3x (no implicit multiplication), and wrap fractions in parentheses ((x+1)/(x−2), not x+1/x−2 which the parser reads as x + 1/x − 2 by operator precedence).
A few common causes. Division by zero (1/x at x = 0, tan(x) at x = π/2). Logarithm of a non-positive number (ln(0), ln(−2)) since the page works in real arithmetic. Square root of a negative number. Improper integrals that diverge: ∫₁^∞ 1/x dx is +∞, and the page reports it as such or as "undefined." For piecewise issues, plot f first to see asymptotes and discontinuities. If the integral crosses a vertical asymptote, split it at the singularity and check convergence on each piece.
Yes. Use the order parameter or differentiate the result repeatedly. f''(x) is the derivative of f'(x); f'''(x) is the derivative of f''. For x⁴: f'(x) = 4x³, f''(x) = 12x², f'''(x) = 24x, f''''(x) = 24. Practical uses: second derivative test for max/min (f'(a) = 0 and f''(a) > 0 means local minimum), concavity (f'' > 0 concave up), inflection points (f'' = 0 and changes sign), and Taylor series coefficients (f^(n)(a)/n!).
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