Multivariable Chain Rules and Implicit Surfaces

Track every dependency through a composition and use a nonzero partial to solve locally for a variable.

Builds on Gradients and Directional Derivatives

The bigger question: How does a surface change in different directions?

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Add contributions from every changing input

If z=f(x(t),y(t))z=f(x(t),y(t)) and the relevant maps are differentiable, then

dzdt=fxx′+fyy′.\frac{dz}{dt}=f_x x'+f_y y'.

If x,yx,y depend on variables u,vu,v, apply the same rule separately for each input: zu=fxxu+fyyuz_u=f_xx_u+f_yy_u. Evaluate outer derivatives at the inner point. The Jacobian matrix organizes these relations as D(f∘g)=Df(g)DgD(f\circ g)=Df(g)Dg, with compatible matrix dimensions.

Visual guide

VISUAL GUIDEA tangent plane touches a spherical surface
This cross-section fixes y = 2 in the sphere x² + y² + z² = 9. At (x, z) = (1, 2), implicit differentiation gives dz/dx = −1/2. The tangent line shown is the corresponding slice of the 3D tangent plane.-2.7-2.7-1.35-1.35001.351.352.72.7xz
  • Sphere slice x² + z² = 5
  • Tangent z = 2 − (x − 1)/2
This cross-section fixes y = 2 in the sphere x² + y² + z² = 9. At (x, z) = (1, 2), implicit differentiation gives dz/dx = −1/2. The tangent line shown is the corresponding slice of the 3D tangent plane.

Worked example: a path across a surface

Let z=x2yz=x^2y, with x=t2x=t^2 and y=sin⁡ty=\sin t. The chain rule gives

z′=2xy(2t)+x2cos⁡t=4t3sin⁡t+t4cos⁡t.z'=2xy(2t)+x^2\cos t=4t^3\sin t+t^4\cos t.

Substituting first gives z=t4sin⁡tz=t^4\sin t, whose ordinary derivative agrees. Agreement between the two routes is a useful check when both are manageable.

For a temperature field T(x,y,t)T(x,y,t) experienced by a moving sensor, the total rate is dT/dt=Txx′+Tyy′+TtdT/dt=T_xx'+T_yy'+T_t. The explicit time term is separate from motion through spatial variation.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

For f(x,y)=xy with x=t and y=t², what is df/dt?

Hint 1 · Find a starting point

Both x and y depend on t.

Hint 2 · Take the next step

Use fₓx′+fᵧy′, or first substitute f=t³.

Show the reasoning

Answer: 3t²

t²×1+t×2t=3t², matching the derivative of t³.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: solve for a surface height

Suppose F(x,y,z)=x2+y2+z2−9=0F(x,y,z)=x^2+y^2+z^2-9=0. Where Fz=2z≠0F_z=2z\ne0, the implicit function theorem allows a local differentiable height z(x,y)z(x,y), with

zx=−Fx/Fz=−x/z,z_x=-F_x/F_z=-x/z, zy=−Fy/Fz=−y/z.z_y=-F_y/F_z=-y/z.

At (1,2,2)(1,2,2), the slopes are −1/2-1/2 and −1-1. The tangent plane is z−2=−(x−1)/2−(y−2)z-2=-(x-1)/2-(y-2), equivalently x+2y+2z=9x+2y+2z=9.

At equator points where z=0z=0, this height formula fails because the sphere is vertical relative to the xyxy-plane. The sphere itself is still smooth there: another nonzero partial can provide a different local graph. Failure of one coordinate representation is not necessarily a singular surface.

Practice

  1. Find d(xy)/dtd(xy)/dt when x=et,y=t2x=e^t,y=t^2.
  2. Find zxz_x for xz+y+z2=0xz+y+z^2=0 where x+2z≠0x+2z\ne0.
  3. What term is missing from dT/dt=Txx′+Tyy′dT/dt=T_xx'+T_yy' when the field itself changes with time?
Show worked solutions
  1. ett2+2tete^tt^2+2te^t.
  2. Fx=zF_x=z and Fz=x+2zF_z=x+2z, giving zx=−z/(x+2z)z_x=-z/(x+2z).
  3. The explicit partial derivative TtT_t.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.

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