Chain rule for multivariable functions

Chain rule for multivariable functions

Lessons

Notes:


Suppose we have a function y=f(g(x))y=f(g(x)). Then the chain rule is:

y =f(g(x))g(x)y\ = f'(g(x))g'(x)

We can rewrite this using an alternate notation:

dydx=dfdgdgdx \frac{dy}{dx} = \frac{df}{dg} \frac{dg}{dx}

Now if we were to change gxg \to x and xtx\to t, then we have the chain rule to be:

dydt=dfdxdxdt\frac{dy}{dt} = \frac{df}{dx} \frac{dx}{dt}

Why do we want this alternate notation? Because it relates to the chain rule for 2 variable functions.

1st Case of Chain Rule for 2 Variable Functions
Suppose we have z=f(x,y)z=f(x,y), x=g(t)x=g(t), and y=h(t)y=h(t), then the chain rule (derivative of zz in respect to tt) is:

dzdt=dfdxdxdt+dfdydydt\frac{dz}{dt} = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt}


2nd Case of Chain Rule for 2 Variable Functions
Suppose we have z=f(x,y)z=f(x,y),x=g(s,t) x=g(s, t),y=h(s,t) y=h(s, t), then there are 2 chain rules.
The derivative of zz in respect to ss is:

dzdx=dfdxdxdx+dfdydyds\frac{dz}{dx} = \frac{df}{dx} \frac{dx}{dx} + \frac{df}{dy} \frac{dy}{ds}

The derivative of zz in respect to tt is:

dzdt=dfdxdxdt+dfdydydt \frac{dz}{dt} = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt}


Using a Tree Diagram for Chain Rule
Tree diagrams are very useful when finding the chain rule for multivariable functions with more than 2 variables.
For example, suppose we have w=f(x,y,z),x=g(s,t,r),y=h(s,t,r)w=f(x,y,z), x=g(s,t,r), y=h(s,t,r) and z=i(s,t,r)z=i(s,t,r), and we want to find dwdt\frac{dw}{dt}.
We can write the tree diagram below like this:
chain rule tree diagram
Then we will multiply all the connected derivatives, and sum them up to have:

dwdt=dfdxdxdt+dfdydydt+dfdzdzdt\frac{dw}{dt} = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt} + \frac{df}{dz} \frac{dz}{dt}

  • Introduction
    Chain Rule for Multivariable Functions Overview:
    a)
    A Review of Chain Rule
    • What is the chain rule?
    • h(x)=f(g(x))h=f(g(x))g(x)h(x)=f(g(x))\to h' =f'(g(x))g'(x)
    • alternate notation: dtdt=dfdxdxdt\frac{dt}{dt} = \frac{df}{dx} \frac{dx}{dt}

    b)
    1st Case of Chain Rule for 2 Variable Functions
    • z=f(x,y),x=g(t),y=h(t)z=f(x,y), x=g(t), y=h(t)
    • Then dzdt=dfdxdxdt+dfdydydt \frac{dz}{dt} = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt}
    • An example

    c)
    2nd Case of Chain Rule for 2 Variable Functions
    • z=f(x,y),x=g(s,t),y=h(s,t)z=f(x,y), x=g(s, t), y=h(s, t)
    • dzdx=dfdxdxdx+dfdydyds\frac{dz}{dx} = \frac{df}{dx} \frac{dx}{dx} + \frac{df}{dy} \frac{dy}{ds}
    • dzdt=dfdxdxdt+dfdydydt \frac{dz}{dt} = \frac{df}{dx} \frac{dx}{dt} + \frac{df}{dy} \frac{dy}{dt}
    • An example

    d)
    Using a Tree Diagram for Chain Rule
    • Easier to find the chain rule for multivariable functions
    • What it looks like
    • Couple Examples


  • 1.
    Finding the Chain Rule for 2 Variable Functions
    Given that z=sin(xy),x=t3,y=1tz=\sin (xy), x= t^3, y=1-t, find dzdt \frac{dz}{dt} .

  • 2.
    Given that z=ln(x2+y2),x=s2+t2,y=2s+t z= \ln (x^2+y^2), x=\sqrt{s^2+t^2}, y=2s+t , find dzds\frac{dz}{ds} .

  • 3.
    Finding the Chain Rule for Implicit Functions
    Given that xy2+cos(x2y)=3xy^2+ \cos (x^2 y)=3, find dydx\frac{dy}{dx}.

  • 4.
    Given that exz+ln(xy)=x2y3e^{xz}+ \ln (xy)=x^2y^3, find dzdx\frac{dz}{dx}.

  • 5.
    Using the Tree Diagram to get the Chain Rule
    Suppose w=f(x,y,z),x=g(s,t),y=h(s,t)w=f(x, y, z), x=g(s, t), y=h(s, t), z=i(s,t)z=i(s, t). Use the tree diagram and use the chain rule to find dwdt\frac{dw}{dt}.

  • 6.
    Suppose w=f(x,y,z),x=g(r,s,t),y=h(r,s,t),z=h(r,s,t),r=a(q).w=f(x, y, z), x=g(r, s, t), y=h(r, s, t), z=h(r, s, t), r=a(q). Use the tree diagram and use the chain rule to find dwdq\frac{dw}{dq}.