### Multivariable Chain Rule (Example 2)

You use the multivariable chain rule when you have functions that are functions of other functions, basically it has equations that look like they could be substituted into each other. Please see the solved example:

Find $$\frac{{dz}}{{ds}}$$ and $$\frac{{dz}}{{dt}}$$ using the chain rule, where $$z = {x^2}{y^3}$$, $$x = s\cos (t)$$, and $$y = s\sin (t)$$.

See how you could technically substitute the equations into each other and then find dz/dt that way? But it's actually a lot easier to use the multivariable chain rule. The best way to do it is map out how the variables are related to each other. z is a function of x, which is a function of s, and z is a function of y, which is a function of s. As well, z is a function of x, which is a function of t, and z is a function of y, which is a function of t.

First let's find $$\frac{{dz}}{{dt}}$$.

$\frac{{dz}}{{ds}} = \frac{{\partial z}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial s}} + \frac{{\partial z}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial s}}$

We find all the required partial derivatives and substitute them into the equation above.

$\frac{{\partial z}}{{\partial x}} = 2x{y^3}$

$\frac{{\partial x}}{{\partial s}} = \cos (t)$

$\frac{{\partial z}}{{\partial y}} = 3{x^2}{y^2}$

$\frac{{\partial z}}{{\partial s}} = \sin (t)$

Substituting these in,

$\frac{{dz}}{{ds}} = 2x{y^3}\cos (t) + 3{x^2}{y^2}\sin (t)$

And now let's find dz/dt. Now we use,

$\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial z}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}$

Find these partial derivatives and substitute them in.

$\frac{{\partial z}}{{\partial x}} = 2x{y^3}$

$\frac{{\partial x}}{{\partial t}} - s\sin (t)$

$\frac{{\partial z}}{{\partial y}} = 3{x^2}{y^2}$

$\frac{{\partial y}}{{\partial t}} = s\cos (t)$

Substituting these in we obtain:

$\frac{{dz}}{{dt}} = - 2sx{y^3}\sin (t) + 3s{x^2}{y^2}\cos (t)$

Depending on your instructor's preference, you may need to substitute x and y provided in the question to get dz/dx and dz/ds as a function of s and t only.

Please see our other examples on partial derivatives, including chain rule, product rule, quotient rule, and multivariable implicit differentiation:

Partial Derivatives Examples:

Partial Derivatives (Ex1)

Partial Derivatives (Ex2)

Partial Derivatives (Ex3)

Partial Derivatives (Ex4)

Multivariable Chain Rule Examples:

Multivariable Chain Rule (Ex1)

Implicit Multivariable Differentiation:

Multivariable Implicit Differentiation

Return to the multivariable differentiation hub page:

Partial Derivatives, Multivariable Differentiation