Partial Derivatives & Chain Rule

Before diving into the topics, make sure you're solid on the building blocks.

Three quick cards: partial derivatives as slopes, the tangent plane, and the multivariable chain rule.

Partial Derivatives as Slopes

For $z = f(x,y)$, the partial derivative $\frac{\partial f}{\partial x}$ is the slope of the surface when you slice along a plane of constant $y$, and vice versa.

$\displaystyle \frac{\partial f}{\partial x}\bigg|_{(a,b)} = \lim_{h\to 0}\frac{f(a+h,\,b) - f(a,\,b)}{h}$

The surface below is $z = x^2 - y^2$ (a saddle). Toggle the slice planes and move the slider to see how the slope changes.

Drag to rotate

Position: 0.50

x-slice (y=0.50): slope = ∂f/∂x = 2x = 1.00
y-slice (x=0.50): slope = ∂f/∂y = −2y = −1.00

Total Differential & Tangent Plane

At any point $(a,b)$ on the surface $z = f(x,y)$, the tangent plane is the best linear approximation:

$\displaystyle dz = \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy \qquad\Longleftrightarrow\qquad z - f(a,b) = f_x(a,b)(x-a) + f_y(a,b)(y-b)$

Move the point around on $z = x^2 - y^2$ and watch the semi-transparent tangent plane tilt.

Drag to rotate

x = 0.80 y = 0.40

Point: (0.80, 0.40) → z = 0.48
f_x = 1.60, f_y = −0.80
Tangent plane: z = 0.48 + 1.60(x−0.80) − 0.80(y−0.40)

The Chain Rule

If $z = f(x,y)$ and $x = x(t)$, $y = y(t)$, then the derivative of $z$ with respect to $t$ decomposes into two contribution paths:

$\displaystyle \frac{dz}{dt} = \underbrace{\frac{\partial z}{\partial x}\cdot\frac{dx}{dt}}_{\text{x-path}} + \underbrace{\frac{\partial z}{\partial y}\cdot\frac{dy}{dt}}_{\text{y-path}}$

Example: $z = x^2 - y^2$, $x(t) = \cos t$, $y(t) = \sin t$.

t = 0.79

x = 0.71

y = 0.71

z = 0.00

∂z/∂x = 1.41

dx/dt = −0.71

∂z/∂y = −1.41

dy/dt = 0.71

dz/dt = −2.00

Notice that when $t = \pi/4$, the x-path contributes $-1.00$ and the y-path contributes $-1.00$, summing to $dz/dt = -2.00$. Slide $t$ to see both terms change.