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Wize University Calculus 3 Textbook > Partial Derivatives

Chain Rules for Functions with Several Variables & Implicit Differentiation

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Chain Rule

From Single Variable Calculus, the Chain Rule is defined as:
F(x)=f(g(x))      →      F′(x)=f′(g(x))⋅g′(x)F(x)=f(g(x))~~~~~~{\color{red}\rightarrow}~~~~~~F'(x)=f'(g(x))\cdot g'(x)

Or, if y=f(x) and x=g(t), y=f(x)~\text{and}~x=g(t),~then:
∂y∂t=∂y∂x⋅∂x∂t\frac{\partial{y}}{\partial{t}}=\frac{\partial{y}}{\partial{x}}\cdot \frac{\partial{x}}{\partial{t}}

In Multivariable Calculus, there are 2 cases to the Chain Rule.

Chain Rule

  1. Case 1:
  • If z=f(x,y)z=f(x,y), x=g(t)x=g(t), and y=h(t)y=h(t), then:
dzdt=∂f∂x⋅dxdt+∂f∂y⋅dydt\boxed{\dfrac{dz}{dt}=\dfrac{\partial f}{\partial x}\cdot{}\dfrac{dx}{dt}+\dfrac{\partial f}{\partial y}\cdot{}\dfrac{dy}{dt}}
  1. Case 2:
  • If z=f(x,y)z=f(x,y), x=g(t,s)x=g(t,s), y=h(t,s)y=h(t,s), then:
∂z∂t=∂f∂x∂x∂t+∂f∂y∂y∂t∂z∂s=∂f∂x∂x∂s+∂f∂y∂y∂s\boxed{\begin{array}{rcl} \begin{array}{l} \dfrac{\partial z}{\partial t}&=&\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial t}\\\\ \dfrac{\partial z}{\partial s}&=&\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial s}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial s} \end{array} \end{array}}

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Chain Rule & Implicit Differentiation

One of the applications of the chain rule is implicit differentiation.

Case 1

  • Let F(x,y)=0F(x,y)=0
  • yy is a function of xx in the form of y=f(x)y=f(x)

Applying the chain rule with respect to xx, we will have:

∂F∂x⋅∂x∂x+∂F∂y⋅∂y∂x=0\dfrac{\partial F}{\partial x}\cdot\dfrac{\partial x}{\partial x}+\dfrac{\partial F}{\partial y}\cdot\dfrac{\partial y}{\partial x}=0
Then,
dydx=−(∂F∂x)(∂F∂y)=−FxFy\boxed{\dfrac{dy}{dx}=-\dfrac{\Bigg(\dfrac{\partial F}{\partial x}\Bigg)}{\Bigg(\dfrac{\partial F}{\partial y}\Bigg)}=-\dfrac{F_x}{F_y}}

Case 2

  • Let F(x,y,z)=0F(x,y,z)=0
  • zz is a function of xx and yy in the form z=f(x,y)z=f(x,y)
In this case, it can be proved that the partial derivative of zz with respect to xx and yy are:

∂z∂x=−(∂F∂x)(∂F∂z)=−FxFz∂z∂y=−(∂F∂y)(∂F∂z)=−FyFz\boxed{\begin{array}{l} \dfrac{\partial z}{\partial x}=-\dfrac{\Bigg(\dfrac{\partial F}{\partial x}\Bigg)}{\Bigg(\dfrac{\partial F}{\partial z}\Bigg)}=-\dfrac{F_x}{F_z}\\\\ \dfrac{\partial z}{\partial y}=-\dfrac{\Bigg(\dfrac{\partial F}{\partial y}\Bigg)}{\Bigg(\dfrac{\partial F}{\partial z}\Bigg)}=-\dfrac{F_y}{F_z} \end{array}}

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Example

If w=f(x,y,z)w=f(x,y,z), x=g(u,v)x=g(u,v), y=h(u,v)y=h(u,v), and z=k(u,v)z=k(u,v), find ∂w∂u\dfrac{\partial w}{\partial u} and ∂w∂v\dfrac{\partial w}{\partial v}


∂w∂u=∂w∂x⋅∂x∂u+∂w∂y⋅∂y∂u+∂w∂z⋅∂z∂u(1)(3)(5)∂w∂v=∂w∂x⋅∂x∂v+∂w∂y⋅∂y∂v+∂w∂z⋅∂z∂v(2)(4)(6)\begin{array}{ccccccc} \dfrac{\partial w}{\partial u}&=&\dfrac{\partial w}{\partial x}\cdot\dfrac{\partial x}{\partial u}&+&\dfrac{\partial w}{\partial y}\cdot\dfrac{\partial y}{\partial u}&+&\dfrac{\partial w}{\partial z}\cdot\dfrac{\partial z}{\partial u}\\\\ &&(1)&&(3)&&(5)\\\\ \dfrac{\partial w}{\partial v}&=&\dfrac{\partial w}{\partial x}\cdot\dfrac{\partial x}{\partial v}&+& \dfrac{\partial w}{\partial y}\cdot\dfrac{\partial y}{\partial v}&+&\dfrac{\partial w}{\partial z}\cdot\dfrac{\partial z}{\partial v}\\\\ &&(2)&&(4)&&(6) \end{array}


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Example

Calculate dydx\dfrac{dy}{dx} if x3cos⁡(y)+y4=0x^3\cos(y)+y^4=0

x3cos⁡(y)+y4=0x^3\cos(y)+y^4=0
dydx=−∂F∂x∂F∂y\frac{dy}{dx}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}
when F(x,y)=x3cos⁡(y)+y4=0F(x,y)=x^3\cos(y)+y^4=0
Fx=3x2cos⁡(y),Fy=−x3sin⁡(y)+4y3F_x=3x^2\cos(y),\qquad\qquad F_y=-x^3\sin(y)+4y^3

⇒ dydx=−3x2cos⁡(y)−x3sin⁡(y)+4y3\Rightarrow\ \frac{dy}{dx}=-\frac{3x^2\cos(y)}{-x^3\sin(y)+4y^3}

Practice


Consider the surface F(x, y, z)=ex2+y2+z2−(x2+y2+z2)=0F(x,~y,~z)=e^{x^2+y^2+z^2}-(x^2+y^2+z^2)=0.

Find ∂z∂x & ∂z∂y\dfrac{\partial{z}}{\partial{x}}~\&~\dfrac{\partial{z}}{\partial{y}}.


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Practice

Find ∂z∂t \dfrac{\partial{z}}{\partial{t}}~for z=xexyz=xe^{\frac{x}{y}}, where x=t3x=t^3 and y=ty=\sqrt{t}.