Partial Derivative Calculator

Q: What are second-order and mixed partials?

f_{xx} differentiates by x twice; f_{xy} differentiates by x then y. For smooth functions f_{xy} = f_{yx} (Clairaut’s theorem). The calculator shows all of them so you can check.

Q: What is the gradient?

The vector of all first partials, \nabla f = \left(f_x, f_y\right). It points in the direction of steepest increase and its length is that steepest slope.

Q: What does the graph show?

A slice of the surface: f(x, y_0) with y frozen at your chosen value. The slope of the pink tangent on that slice is exactly \frac{\partial f}{\partial x} at the point.

Differentiate with respect to one variable while holding the others still. Get every first and second partial, the gradient at a point, and a slice of the surface that makes ∂f/∂x visible.

Function f(x, y) (z allowed too)

Reads as

x^{2} y^{3} + \sin\left(x y\right)

Try:

With respect to:

Point x =y =

Answer

\frac{\partial}{\partial x}\left[x^{2} y^{3} + \sin\left(x y\right)\right] = 2 x y^{3} + y \cos\left(x y\right)

$\nabla f(1, 2) = \left(15.1677,\ 11.5839\right)$

f(x, 2)

Slice at y = 2. Tangent slope = ∂f/∂x = 15.168.

Steps for ∂f/∂x (other variables act as constants)

Sum rule
Differentiate term by term.
$\frac{d}{dx}\left[x^{2} y^{3} + \sin\left(x y\right)\right] = \frac{d}{dx}\left[x^{2} y^{3}\right] + \frac{d}{dx}\left[\sin\left(x y\right)\right]$
Constant multiple rule
Pull the constant factor out front.
$\frac{d}{dx}\left[x^{2} y^{3}\right] = y^{3} \cdot \frac{d}{dx}\left[x^{2}\right]$
Power rule
d/dx[xⁿ] = n·xⁿ⁻¹.
$\frac{d}{dx}\left[x^{2}\right] = 2 x$
Constant multiple rule
Pull the constant factor out front.
$\frac{d}{dx}\left[x y\right] = y \cdot \frac{d}{dx}\left[x\right]$
Trig rule + chain rule
Outer function $\sin\left(□\right)$ , inner u = $x y$ ; multiply the outer derivative by u′.
$\frac{d}{dx}\left[\sin\left(x y\right)\right] = \cos\left(x y\right) \cdot \frac{d}{dx}\left[x y\right]$

All partial derivatives

f_{x} = 2 x y^{3} + y \cos\left(x y\right)

f_{y} = 3 x^{2} y^{2} + x \cos\left(x y\right)

f_{xx} = 2 y^{3} - y^{2} \sin\left(x y\right)

f_{xy} = 6 x y^{2} - x y \sin\left(x y\right) + \cos\left(x y\right)

f_{yx} = 6 x y^{2} - x y \sin\left(x y\right) + \cos\left(x y\right)

f_{yy} = 6 y x^{2} - x^{2} \sin\left(x y\right)

Still stuck? Ask the tutor

An AI tutor reads the problem and steps above and explains them in plain words. It can make mistakes, so check it against the worked steps.

Freeze, then differentiate

To find $\frac{\partial f}{\partial x}$ , pretend every other letter is a fixed number and differentiate as usual. Geometrically, you are slicing the surface $z = f(x,y)$ with a plane $y = y_0$ and measuring the slope of the curve you cut out. That is the graph above: drag along it and the tangent’s slope is the partial derivative.

Example

For $f(x,y) = x^2y^3$ : $f_x = 2xy^3$ (treat $y^3$ as a constant factor) and $f_y = 3x^2y^2$ (treat $x^2$ as a constant factor). The mixed partials agree: $f_{xy} = f_{yx} = 6xy^2$ .

Questions students ask

What is a partial derivative?

The rate of change of a multivariable function in one direction, holding every other variable fixed. $\frac{\partial f}{\partial x}$ treats $y$ as a constant; $\frac{\partial f}{\partial y}$ treats $x$ as a constant.

How is it different from an ordinary derivative?

The rules are identical; the only change is that other variables behave like numbers. For $f = x^2y^3$ , $\frac{\partial f}{\partial x} = 2xy^3$ exactly as if $y^3$ were 5.

What are second-order and mixed partials?

$f_{xx}$ differentiates by $x$ twice; $f_{xy}$ differentiates by $x$ then $y$ . For smooth functions $f_{xy} = f_{yx}$ (Clairaut’s theorem). The calculator shows all of them so you can check.

What is the gradient?

The vector of all first partials, $\nabla f = \left(f_x, f_y\right)$ . It points in the direction of steepest increase and its length is that steepest slope.

What does the graph show?

A slice of the surface: $f(x, y_0)$ with $y$ frozen at your chosen value. The slope of the pink tangent on that slice is exactly $\frac{\partial f}{\partial x}$ at the point.