Multivariable Chain Rule

Topics:

• Be able to compute partial derivatives with the various versions of the multivariate chain rule.
• Be able to compare your answer with the direct method of computing the partial derivatives.

Given a multivariable function $f(x,y)$ and two single variable functions $x(t)$ and $y(t)$, the multivariable chain rule says:

$\underset{\text{Derivative of composition function}}{\underbrace{\frac{d}{dt}f(x(t),y(t))}}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$

Written in vector notation, where $\vec{v}(t)=\left[\begin{array}{c}x(t)\\ y(t)\end{array}\right]$, this rule as a very elegant form in terms of the [[The Gradient and Directional Derivative]|the gradient]] of $f$ and the [vector derivative] of $\vec{v}(t)$:

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$\underset{\text{Derivative of composition function}}{\underbrace{\frac{d}{dt}f(\vec{v}(t))}}=\stackrel{\text{Dot product of vectors}}{\overbrace{\nabla f\cdot {\vec{v}}^{\prime }(t)}}$

Whereas the single variable chain rule looks like this:

$\frac{d}{dt}f(g(t))=\frac{df}{dg}\frac{dg}{dt}=f^{\prime }(g(t))g^{\prime }(t)$

What if instead of taking in a one-dimensional input, $t$, the function $f$ took in a two-dimensional input, $(x,y)$?

Then it wouldn’t make sense to have a scalar valued function like $g(t)$, instead we can have two separate scalar valued functions and plug them in as coordinates for $f$.

$$\begin{array}{rrcl}\text{One final output}& & f(x(t),y(t))& \\ & & & \\ & \nearrow & & \nwarrow \\ & & & \\ \text{Two intermediate outputs}& x(t)& & y(t)\\ & & & \\ & \nwarrow & & \nearrow \\ & & & \\ \text{One input}& & t& \end{array}$$

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Example from Webwork Week 5 Problem 5 Consider the surface $x^8{z}^3+\sin (y^3{z}^3)+8=0$. Use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.

$$\begin{array}{rl}\frac{\partial f}{\partial x}[x^8{z}^3+\sin (y^3{z}^3)+8]& =8x^7{z}^3+x^8(3z^2\frac{\partial z}{\partial x})+\cos (y^3{z}^3)(y^3)(3z^2\frac{\partial z}{\partial x})\\ \frac{\partial f}{\partial y}[...]& =x^8(3z^2\frac{\partial z}{\partial y})+\cos (y^3{z}^3)(3y^2{z}^3+y^3(3z^2\frac{\partial z}{\partial y}))\\ \frac{\partial f}{\partial z}[...]& =3z^2{x}^8+3y^3{z}^2\cos (y^3{z}^3)\\ & \\ \frac{\partial z}{\partial x}& =\frac{-f_x}{f_z}\\ & =\frac{-8x^7{z}^3}{3z^2(x^8+y^3\cos (y^3{z}^3))}\\ & \\ \frac{\partial z}{\partial y}& =\frac{-f_y}{f_z}\\ & =\frac{-y^2{z}^3\cos (y^3{z}^3)}{z^2(x^8+y^3\cos (y^3{z}^3))}\end{array}$$

Webwork Week 5 Problem 6 Let $f(x,y,z)=x^{3}y+z^2$ and $x=s^{3}t$, $y=s^{2}t$, and $z=s^2{t}^2$.

$$\begin{array}{rl}\frac{\partial f}{\partial x}& =3x^{2}y\\ \frac{\partial f}{\partial y}& =x^3\\ \frac{\partial f}{\partial z}& =2z\\ & \\ \frac{\partial x}{\partial s}& =3s^{2}t\\ \frac{\partial y}{\partial s}& =2st\\ \frac{\partial z}{\partial s}& =2s{t}^2\\ & \\ \frac{\partial f}{\partial s}& =\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial s}\\ & \\ & =(3x^{2}y)(3s^{2}t)+(x^3)(2st)+(2z)(2s{t}^2)\\ & =(3(s^{3}t{)}^2(s^{2}t))(3s^{2}t)+((s^{3}t{)}^3)(2st)+(2(s^2{t}^2))(2s{t}^2)\\ & =3(s^8{t}^3)(3s^{2}t)+2s^{10}{t}^4+4(s^3{t}^4)\\ & =9(s^{10}{t}^4)+2s^{10}{t}^4+4(s^3{t}^4)\end{array}$$

1/7/25