Wize University Calculus 2 Textbook > Multivariable Functions

Chain Rule

Gradient and the directional derivativeNext Section

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Chain Rule for Partial Derivatives
Disclaimer: At timestamp 4:19, the equation should read [2s-rt] instead of e^[2s-rt].

In many cases, given a function of two or more variables, each variables could be a function of another variable.

Example
If f(x,y)=2xy+x2−y2f\left(x,y\right)=2xy+x^2-y^2f(x,y)=2xy+x2−y2 where x(t)=2t+1x\left(t\right)=2t+1x(t)=2t+1 and y(t)=cos⁡ty\left(t\right)=\cos ty(t)=cost, evaluate f(x,y)f\left(x,y\right)f(x,y) at t=0t=0t=0.

x(0)=2(0)+1=1x\left(0\right)=2\left(0\right)+1=1x(0)=2(0)+1=1 and y(0)=cos⁡0=1y\left(0\right)=\cos0=1y(0)=cos0=1
So, f(1,1)=2(1)(1)+12−12=2f\left(1,1\right)=2\left(1\right)\left(1\right)+1^2-1^2=2f(1,1)=2(1)(1)+12−12=2

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Chain Rule
Suppose that z=f(x,y)z=f\left(x,y\right)z=f(x,y) is a differentiable function where x=g(t)x=g\left(t\right)x=g(t) and y=h(t)y=h\left(t\right)y=h(t), then
dzdt=∂z∂x dxdt+∂z∂y dydt\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\ \frac{dx}{dt}+\frac{\partial z}{\partial y}\ \frac{dy}{dt}dtdz​=∂x∂z​ dtdx​+∂y∂z​ dtdy​

Suppose that z=f(x,y)z=f(x,y)z=f(x,y) is a differentiable function where x=g(s,t)x=g(s,t)x=g(s,t) and y=h(s,t)y=h(s,t)y=h(s,t), then
∂z∂s=∂z∂x ∂x∂s+∂z∂y ∂y∂s\displaystyle\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial s}∂s∂z​=∂x∂z​ ∂s∂x​+∂y∂z​ ∂s∂y​  and  ∂z∂t=∂z∂x ∂x∂t+∂z∂y ∂y∂t\displaystyle\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial t}∂t∂z​=∂x∂z​ ∂t∂x​+∂y∂z​ ∂t∂y​

Strategy
Use a “tree diagram” to list all the variables and how they depend on one another
Consider all possible paths from the top of the tree diagram to the variable you want to differentiate with respect to
Multiply along the path
Add the results of different paths
Which symbol to use?
When differentiating a variable that depend on 2 or more variables, use the ∂...∂...\frac{\partial...}{\partial...}∂...∂...​ symbol
When differentiating a variable that only depends on 1 variable, use the d...d...\frac{d...}{d...}d...d...​ symbol

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Example
If z=exy+3x2y2z=e^xy+3x^2y^2z=exy+3x2y2, x=2s−rtx=2s-rtx=2s−rt and y=s2ty=s^2ty=s2t, find ∂z∂r, ∂z∂s, and  ∂z∂t\frac{\partial z}{\partial r},\ \frac{\partial z}{\partial s},\ \text{and }\ \frac{\partial z}{\partial t}∂r∂z​, ∂s∂z​, and  ∂t∂z​

∂z∂r=∂z∂x ∂x∂r+∂z∂y ∂y∂r\displaystyle\frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial r}∂r∂z​=∂x∂z​ ∂r∂x​+∂y∂z​ ∂r∂y​
∂z∂r=(yex+6y2x)(−t)\displaystyle\frac{\partial z}{\partial r}=\left(ye^x+6y^2x\right)\left(-t\right)∂r∂z​=(yex+6y2x)(−t)
∂z∂r=−t(s2te2s−rt+6s4t2(2s−rt))\displaystyle\frac{\partial z}{\partial r}=-t\left(s^2te^{2s-rt}+6s^4t^2\left(2s-rt\right)\right)∂r∂z​=−t(s2te2s−rt+6s4t2(2s−rt))

∂z∂s=∂z∂x ∂x∂s+∂z∂y ∂y∂s\displaystyle\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial s}∂s∂z​=∂x∂z​ ∂s∂x​+∂y∂z​ ∂s∂y​
∂z∂s=(yex+6y2x)(2)+(ex+6x2y)(2st)\displaystyle\frac{\partial z}{\partial s}=\left(ye^x+6y^2x\right)\left(2\right)+\left(e^x+6x^2y\right)\left(2st\right)∂s∂z​=(yex+6y2x)(2)+(ex+6x2y)(2st)
∂z∂s=2(s2te2s−rt+6s4t2(2s−rt))+(2st)(e2s−rt+6(2s−rt)2s2t)\displaystyle\frac{\partial z}{\partial s}=2\left(s^2te^{2s-rt}+6s^4t^2\left(2s-rt\right)\right)+\left(2st\right)\left(e^{2s-rt}+6\left(2s-rt\right)^2s^2t\right)∂s∂z​=2(s2te2s−rt+6s4t2(2s−rt))+(2st)(e2s−rt+6(2s−rt)2s2t)

∂z∂t=∂z∂x ∂x∂t+∂z∂y ∂y∂t\displaystyle\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial t}∂t∂z​=∂x∂z​ ∂t∂x​+∂y∂z​ ∂t∂y​
∂z∂s=(yex+6y2x)(−r)+(ex+6x2y)(s2)\frac{\partial z}{\partial s}=\left(ye^x+6y^2x\right)\left(-r\right)+\left(e^x+6x^2y\right)\left(s^2\right)∂s∂z​=(yex+6y2x)(−r)+(ex+6x2y)(s2)
∂z∂s=−r(s2te2s−rt+6s4t2(2s−rt))+s2(e2s−rt+6(2s−rt)2s2t)\displaystyle\frac{\partial z}{\partial s}=-r\left(s^2te^{2s-rt}+6s^4t^2\left(2s-rt\right)\right)+s^2\left(e^{2s-rt}+6\left(2s-rt\right)^2s^2t\right)∂s∂z​=−r(s2te2s−rt+6s4t2(2s−rt))+s2(e2s−rt+6(2s−rt)2s2t)

Practice: Chain Rule
Write the chain rule for ∂z∂u \frac{\partial z}{\partial u}∂u∂z​ if z=f(x,y,u,v), x=x(u,v)z = f (x, y, u, v),\ x = x(u, v)z=f(x,y,u,v), x=x(u,v) and y=y(u,v).y = y(u, v).y=y(u,v). 

∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂u+dzdu\displaystyle \frac{\partial z}{\partial u}=
\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial u}
+
\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial u}+
\frac{d z}{d u}∂u∂z​=∂x∂z​ ∂u∂x​+∂y∂z​ ∂u∂y​+dudz​

∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂u+dzdu+dzdv\displaystyle \frac{\partial z}{\partial u}=
\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial u}
+
\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial u}+
\frac{d z}{d u}+\frac{dz}{dv}∂u∂z​=∂x∂z​ ∂u∂x​+∂y∂z​ ∂u∂y​+dudz​+dvdz​

∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂u\displaystyle \frac{\partial z}{\partial u}=
\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial u}
+
\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial u}∂u∂z​=∂x∂z​ ∂u∂x​+∂y∂z​ ∂u∂y​

∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂u+∂z∂u\displaystyle \frac{\partial z}{\partial u}=
\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial u}
+
\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial u}+
\frac{\partial z}{\partial u}∂u∂z​=∂x∂z​ ∂u∂x​+∂y∂z​ ∂u∂y​+∂u∂z​

∂z∂u=∂z∂x ∂x∂u+∂z∂y ∂y∂u+∂z∂u+∂z∂v\displaystyle \frac{\partial z}{\partial u}=
\frac{\partial z}{\partial x}\ \frac{\partial x}{\partial u}
+
\frac{\partial z}{\partial y}\ \frac{\partial y}{\partial u}+
\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}∂u∂z​=∂x∂z​ ∂u∂x​+∂y∂z​ ∂u∂y​+∂u∂z​+∂v∂z​

I don't know

Practice: Chain Rule
If w=zzyxw = z^z y^xw=zzyx where x=6u+vx=6u+vx=6u+v, y=uvy=uvy=uv and z=tan⁡vz=\tan vz=tanv.

How many terms in ∂w∂u\displaystyle \frac{\partial w}{\partial u}∂u∂w​ has vvv in it?

Practice: Chain Rule
Write the chain rule for ∂z∂r\displaystyle \frac{\partial z}{\partial r}∂r∂z​ and ∂2z∂s ∂r\displaystyle \frac{\partial^2 z}{\partial s\ \partial r}∂s ∂r∂2z​ if z=z(f,g), f=f(r,s), g=g(u,v), u=u(r,s),z = z(f, g),\ f = f(r, s),\ g = g(u,v),\ u  = u(r,  s),z=z(f,g), f=f(r,s), g=g(u,v), u=u(r,s), and v=v(s)v=v(s)v=v(s)

∂z∂r=∂z∂f ∂f∂r+∂z∂g ∂g∂u ∂u∂r\displaystyle \frac{\partial z}{\partial r}=\frac{\partial z}{\partial f}\ \frac{\partial f}{\partial r}+\frac{\partial z}{\partial g}\ \frac{\partial g}{\partial u}\ \frac{\partial u}{\partial r}∂r∂z​=∂f∂z​ ∂r∂f​+∂g∂z​ ∂u∂g​ ∂r∂u​
∂2z∂s ∂r=∂z∂r ∂f∂s+∂z∂r ∂g∂u ∂u∂s\displaystyle \frac{\partial^2z}{\partial s\ \partial r}=
\frac{\partial z}{\partial r}\ \frac{\partial f}{\partial s}
+
\frac{\partial z}{\partial r}\ \frac{\partial g}{\partial u}\ \frac{\partial u}{\partial s}
∂s ∂r∂2z​=∂r∂z​ ∂s∂f​+∂r∂z​ ∂u∂g​ ∂s∂u​

∂z∂r=∂z∂f ∂f∂r+∂z∂g ∂g∂u ∂u∂r\displaystyle \frac{\partial z}{\partial r}=\frac{\partial z}{\partial f}\ \frac{\partial f}{\partial r}+\frac{\partial z}{\partial g}\ \frac{\partial g}{\partial u}\ \frac{\partial u}{\partial r}∂r∂z​=∂f∂z​ ∂r∂f​+∂g∂z​ ∂u∂g​ ∂r∂u​
∂2z∂s ∂r=∂z∂r ∂f∂s+∂z∂r ∂g∂u ∂u∂s+∂z∂r ∂g∂v dvds\displaystyle \frac{\partial^2z}{\partial s\ \partial r}=
\frac{\partial z}{\partial r}\ \frac{\partial f}{\partial s}
+
\frac{\partial z}{\partial r}\ \frac{\partial g}{\partial u}\ \frac{\partial u}{\partial s}
+
\frac{\partial z}{\partial r}\ \frac{\partial g}{\partial v}\ \frac{dv}{ds}∂s ∂r∂2z​=∂r∂z​ ∂s∂f​+∂r∂z​ ∂u∂g​ ∂s∂u​+∂r∂z​ ∂v∂g​ dsdv​

∂z∂r=∂z∂f ∂f∂r+∂z∂g ∂g∂u ∂u∂r\displaystyle \frac{\partial z}{\partial r}=\frac{\partial z}{\partial f}\ \frac{\partial f}{\partial r}+\frac{\partial z}{\partial g}\ \frac{\partial g}{\partial u}\ \frac{\partial u}{\partial r}∂r∂z​=∂f∂z​ ∂r∂f​+∂g∂z​ ∂u∂g​ ∂r∂u​
∂2z∂s ∂r=∂∂f(∂z∂r) ∂f∂s+∂∂g(∂z∂r) ∂g∂u ∂u∂s\displaystyle \frac{\partial^2z}{\partial s\ \partial r}=\frac{\partial}{\partial f}\left(\frac{\partial z}{\partial r}\right)\ \frac{\partial f}{\partial s}+\frac{\partial}{\partial g}\left(\frac{\partial z}{\partial r}\right)\ \frac{\partial g}{\partial u}\ \frac{\partial u}{\partial s}
∂s ∂r∂2z​=∂f∂​(∂r∂z​) ∂s∂f​+∂g∂​(∂r∂z​) ∂u∂g​ ∂s∂u​

∂z∂r=∂z∂f ∂f∂r+∂z∂g ∂g∂u ∂u∂r\displaystyle \frac{\partial z}{\partial r}=\frac{\partial z}{\partial f}\ \frac{\partial f}{\partial r}+\frac{\partial z}{\partial g}\ \frac{\partial g}{\partial u}\ \frac{\partial u}{\partial r}∂r∂z​=∂f∂z​ ∂r∂f​+∂g∂z​ ∂u∂g​ ∂r∂u​
∂2z∂s ∂r=∂∂f(∂z∂r) ∂f∂s+∂∂g(∂z∂r) ∂g∂u ∂u∂s+∂∂g(∂z∂r) ∂g∂v dvds\displaystyle \frac{\partial^2z}{\partial s\ \partial r}=\frac{\partial}{\partial f}\left(\frac{\partial z}{\partial r}\right)\ \frac{\partial f}{\partial s}+\frac{\partial}{\partial g}\left(\frac{\partial z}{\partial r}\right)\ \frac{\partial g}{\partial u}\ \frac{\partial u}{\partial s}+\frac{\partial}{\partial g}\left(\frac{\partial z}{\partial r}\right)\ \frac{\partial g}{\partial v}\ \frac{dv}{ds}∂s ∂r∂2z​=∂f∂​(∂r∂z​) ∂s∂f​+∂g∂​(∂r∂z​) ∂u∂g​ ∂s∂u​+∂g∂​(∂r∂z​) ∂v∂g​ dsdv​

I don't know

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Implicit Differentiation
If we rewrite the function y=f(x)y=f\left(x\right)y=f(x) as F(x,y)=0F\left(x,y\right)=0F(x,y)=0, then
dydx=−FxFy\displaystyle\frac{dy}{dx}=-\frac{F_x}{F_y}dxdy​=−Fy​Fx​​

If we rewrite the function z=f(x,y)z=f(x,y)z=f(x,y) as F(x,y,z)=0F(x,y,z)=0F(x,y,z)=0, then
∂z∂x=−FxFz\displaystyle\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}∂x∂z​=−Fz​Fx​​ and ∂z∂y=−FyFz\displaystyle\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}∂y∂z​=−Fz​Fy​​

Example
If x2+y2z−z3=1−xx^2+y^2z-z^3=1-xx2+y2z−z3=1−x, find ∂z∂x\frac{\partial z}{\partial x}∂x∂z​ and ∂z∂y\frac{\partial z}{\partial y}∂y∂z​.

First we rewrite the function as F(x,y,z)=x2+y2z−z3+x−1=0F\left(x,y,z\right)=x^2+y^2z-z^3+x-1=0F(x,y,z)=x2+y2z−z3+x−1=0
∂z∂x=−FxFz=−2x+1y2−3z2\displaystyle \frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{2x+1}{y^2-3z^2}∂x∂z​=−Fz​Fx​​=−y2−3z22x+1​
∂z∂y=−FyFz=−2zyy2−3z2\displaystyle \frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=-\frac{2zy}{y^2-3z^2}∂y∂z​=−Fz​Fy​​=−y2−3z22zy​

Extra Practice

Practice: Chain Rule
Write the chain rule for ∂z∂u \frac{\partial z}{\partial u}∂u∂z​ if z=f(x,y,u,v), x=x(u,v)z = f (x, y, u, v),\ x = x(u, v)z=f(x,y,u,v), x=x(u,v) and y=y(u,v).y = y(u, v).y=y(u,v). 

Practice: Chain Rule
Write the chain rule for ∂z∂u \frac{\partial z}{\partial u}∂u∂z​ if z=f(x,y,u,v), x=x(u,v)z = f (x, y, u, v),\ x = x(u, v)z=f(x,y,u,v), x=x(u,v) and y=y(u,v).y = y(u, v).y=y(u,v). 

Practice: Chain Rule
Write the chain rule for ∂z∂u \frac{\partial z}{\partial u}∂u∂z​ if z=f(x,y,u,v), x=x(u,v)z = f (x, y, u, v),\ x = x(u, v)z=f(x,y,u,v), x=x(u,v) and y=y(u,v).y = y(u, v).y=y(u,v). 

Practice: Chain Rule
Write the chain rule for ∂z∂u \frac{\partial z}{\partial u}∂u∂z​ if z=f(x,y,u,v), x=x(u,v)z = f (x, y, u, v),\ x = x(u, v)z=f(x,y,u,v), x=x(u,v) and y=y(u,v).y = y(u, v).y=y(u,v). 

Practice: Chain Rule (1)

Consider the surface F(x, y, z)=ex2+y2+z2−(x2+y2+z2)=0.  Find ∂z∂x & ∂z∂y.F(x,~y,~z)=e^{x^2+y^2+z^2}-(x^2+y^2+z^2)=0.~~\text{Find}~\frac{\partial{z}}{\partial{x}}~\&~\frac{\partial{z}}{\partial{y}}.F(x, y, z)=ex2+y2+z2−(x2+y2+z2)=0.  Find ∂x∂z​ & ∂y∂z​.

Practice: Chain Rule

Practice: Chain Rule
If w=zzyxw = z^z y^xw=zzyx where x=6u+vx=6u+vx=6u+v, y=uvy=uvy=uv and z=tan⁡vz=\tan vz=tanv.
How many terms in ∂w∂u\displaystyle \frac{\partial w}{\partial u}∂u∂w​ has vvv in it?

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