standarddeviationcalculator.net

Updated Free · runs in your browser

Math

Multivariable calculus calculator

Enter a function of x and y and choose what to find: the gradient, directional derivative and Hessian at a point, all critical points in a region with their type, or a double integral over a rectangle.

Variables x and y. Write products as x*y or x y.

Any non-zero vector; it is scaled to length 1.

∇f at (2, 1) (9, −3)
∇f(x, y)(3x² − 3y, 3y² − 3x)
f at the point3
Directional derivative along (3, 4)3
Steepest increase |∇f|9.486832981
Hessian determinant D63
Hessian matrix (symbolic)
∂/∂x∂/∂y
∂f/∂x6x−3
∂f/∂y−36y
Hessian matrix at (2, 1)
∂/∂x∂/∂y
∂f/∂x12−3
∂f/∂y−36
Show the working, step by step

  1. The gradient collects the first partial derivatives.

    ∇f = (∂f/∂x, ∂f/∂y) = (3x² − 3y, 3y² − 3x)

  2. Substitute x = 2, y = 1.

    ∇f(2, 1) = (9, −3)

  3. Scale the direction to a unit vector.

    u = (3, 4) ÷ 5 = (0.6, 0.8)

  4. The directional derivative is the dot product of the gradient with u.

    D_u f = 9·0.6 + −3·0.8 = 3

  5. f increases fastest in the direction of ∇f, at rate |∇f|.

    |∇f| = √(9² + −3²) = 9.486832981

  6. The Hessian holds the second partials; its determinant at the point is D = fₓₓfᵧᵧ − fₓᵧ².

    D = 12·6 − (−3)² = 63

The Hessian only classifies a point when the gradient there is zero; switch to "Critical points" to find those.

The formulas

∇f = (fₓ, fᵧ) D_u f = ∇f · u (|u| = 1) H = [ fₓₓ fₓᵧ ; fᵧₓ fᵧᵧ ] D = fₓₓfᵧᵧ − fₓᵧ² ∬ over [a, b] × [c, d] of f dA = ∫ from a to b [ ∫ from c to d of f(x, y) dy ] dx

A worked example

The default function is f(x, y) = x³ − 3xy + y³, a standard example with one saddle and one minimum.

  1. Partial derivatives: fₓ = 3x² − 3y and fᵧ = 3y² − 3x.
  2. At (2, 1): ∇f = (12 − 3, 3 − 6) = (9, −3), and f(2, 1) = 8 − 6 + 1 = 3.
  3. Direction (3, 4) has length 5, so u = (0.6, 0.8) and D_u f = 9·0.6 − 3·0.8 = 3.
  4. Second partials: fₓₓ = 6x, fᵧᵧ = 6y, fₓᵧ = −3. At (2, 1) the Hessian is [12, −3; −3, 6], with D = 72 − 9 = 63.
  5. Critical points: fₓ = 0 gives y = x², then fᵧ = 0 gives x⁴ = x, so x = 0 or 1. At (0, 0), D = 0 − 9 = −9 < 0: a saddle. At (1, 1), D = 36 − 9 = 27 > 0 and fₓₓ = 6 > 0: a local minimum, with f(1, 1) = −1.
  6. Over the unit square [0, 1] × [0, 1]: ∬ (x³ − 3xy + y³) dA = 1/4 − 3/4 + 1/4 = −1/4.

Reading the second-derivative test

DfₓₓCritical point is
> 0> 0local minimum
> 0< 0local maximum
< 0anysaddle point
= 0anynot determined by this test

D is the determinant of the Hessian, the product of its two eigenvalues. Both positive means the surface curves up in every direction (a bowl); both negative, down in every direction (a cap); opposite signs, up one way and down the other (a saddle).

Double integrals

On a rectangle, integrate the inner variable first with the outer one held fixed, then integrate the result. The calculator tries to do this exactly, showing the intermediate function of x, and always computes the value numerically as well with nested adaptive quadrature. Dividing the integral by the rectangle’s area gives the average value of f over it.

Common mistakes

  • Using the direction vector without scaling it to length 1; the result is then |v| times too big.
  • Classifying a point from fₓₓ alone. A positive fₓₓ with D < 0 is still a saddle.
  • Reading "no critical points found" as proof there are none. The search covers only the region given.

Common questions

What does the gradient tell me?

∇f = (∂f/∂x, ∂f/∂y) points in the direction in which f increases fastest, and its length |∇f| is that fastest rate. It is perpendicular to the level curve through the point. For the default function at (2, 1), ∇f = (9, −3) and |∇f| = √90 ≈ 9.487.

How is the directional derivative calculated?

Scale the direction to a unit vector u, then take the dot product with the gradient: D_u f = ∇f · u. For direction (3, 4), u = (0.6, 0.8), so at (2, 1) the rate is 9·0.6 + (−3)·0.8 = 3. The largest possible value is |∇f|, in the gradient’s own direction.

How does the second-derivative test classify a critical point?

Compute D = fₓₓfᵧᵧ − fₓᵧ² at the point. D > 0 with fₓₓ > 0 is a local minimum, D > 0 with fₓₓ < 0 a local maximum, and D < 0 a saddle point. D = 0 gives no answer: x⁴ + y⁴ has a minimum at the origin and x³ + y³ does not, yet both have D = 0 there.

Can the calculator miss a critical point?

Yes. It solves ∇f = 0 numerically with Newton’s method from a grid of starting points inside the search region. Points outside the region are not looked for, and a function with a whole line of critical points (such as x²y²) gives a sample of points along it. Widen the region if you expect more.

Does the order of integration matter in a double integral?

Not for a continuous function on a rectangle: by Fubini’s theorem, integrating over y then x gives the same value as x then y. The order can matter for how easy the inner integral is, and for regions that are not rectangles the limits change when you swap.

Written and reviewed by our editorial team. Last updated . Method and sources: how these numbers are computed.