Chain Rule For Multivariables Tool — Theory
1. Why the multivariable chain rule is needed
In single-variable calculus, the chain rule is used when one function is inside another function.
In multivariable calculus, the same idea appears, but the outer function may depend on several
intermediate variables.
\[
F(t)=f(x(t),y(t)).
\]
Here \(F\) changes because \(x\) changes with \(t\) and because \(y\) changes with \(t\).
2. Chain rule with two intermediate variables
If
\[
F(t)=f(x(t),y(t)),
\]
then the derivative is
\[
\frac{dF}{dt}
=
\frac{\partial f}{\partial x}\frac{dx}{dt}
+
\frac{\partial f}{\partial y}\frac{dy}{dt}.
\]
Short notation:
\[
\frac{dF}{dt}=f_xx_t+f_yy_t.
\]
3. Chain rule with three intermediate variables
If
\[
F(t)=f(x(t),y(t),z(t)),
\]
then
\[
\frac{dF}{dt}
=
f_xx_t+f_yy_t+f_zz_t.
\]
Each term has two factors: an outer sensitivity and an inner rate of change.
4. Two independent variables
Sometimes the intermediate variables depend on two independent variables, for example
\[
x=x(s,t),
\qquad
y=y(s,t),
\qquad
z=z(s,t).
\]
If \(F=f(x,y,z)\), then
\[
\frac{\partial F}{\partial s}
=
f_xx_s+f_yy_s+f_zz_s,
\]
and
\[
\frac{\partial F}{\partial t}
=
f_xx_t+f_yy_t+f_zz_t.
\]
5. Dependency tree idea
A dependency tree shows which variables depend on which other variables.
For
\[
F(t)=f(x(t),y(t)),
\]
the variable \(t\) feeds into both \(x\) and \(y\). Then \(x\) and \(y\) feed into \(f\).
The chain rule adds the contributions from every path from \(t\) to \(F\).
\[
t\to x\to F,
\qquad
t\to y\to F.
\]
6. Example
Let
\[
f(x,y)=x^2+y^2,
\qquad
x=t,
\qquad
y=t^2.
\]
We want to find \(\dfrac{dF}{dt}\), where \(F(t)=f(x(t),y(t))\).
7. Compute the outer partial derivatives
Since
\[
f(x,y)=x^2+y^2,
\]
the partial derivatives are
\[
f_x=2x,
\qquad
f_y=2y.
\]
8. Compute the inner derivatives
Since
\[
x=t,
\qquad
y=t^2,
\]
we get
\[
\frac{dx}{dt}=1,
\qquad
\frac{dy}{dt}=2t.
\]
9. Apply the chain rule
The chain rule says
\[
\frac{dF}{dt}
=
f_x\frac{dx}{dt}
+
f_y\frac{dy}{dt}.
\]
Substitute the pieces:
\[
\frac{dF}{dt}
=
(2x)(1)+(2y)(2t).
\]
Now substitute \(x=t\) and \(y=t^2\):
\[
\frac{dF}{dt}
=
2t+4t^3.
\]
10. Direct-composition check
We can also substitute first:
\[
F(t)
=
f(t,t^2)
=
t^2+(t^2)^2
=
t^2+t^4.
\]
Differentiate directly:
\[
F'(t)=2t+4t^3.
\]
This agrees with the multivariable chain rule result.
12. Common mistakes
- Forgetting one path: every path from the independent variable to the final function contributes a term.
- Forgetting substitution: after computing \(f_x\), \(f_y\), and \(f_z\), substitute the inner functions.
- Mixing variables: when differentiating with respect to \(s\), hold \(t\) constant, and vice versa.
- Using ordinary derivatives for the outer function: the outer function uses partial derivatives because it has several inputs.
- Ignoring the dependency tree: the tree helps organize all terms correctly.
