Total Derivatives – Theory and Solved Problems
Definition
If \(z=f(x,y)\) where \(x\) and \(y\) depend on parameter \(t\), then the total derivative is
\[
\frac{dz}{dt} =
\frac{\partial z}{\partial x}\frac{dx}{dt} +
\frac{\partial z}{\partial y}\frac{dy}{dt}
\]
This formula is known as the **Chain Rule for multivariable functions**.
Visual Understanding of the Chain Rule
In problems involving total derivatives , variables often depend on other variables.
The Chain Rule shows how a change in one variable affects another through intermediate variables.
The following diagrams illustrate this dependency structure.
Diagram 1: Basic Chain Rule Structure
z
x
y
t
This diagram represents:
\[
z = f(x,y), \quad x = x(t), \quad y = y(t)
\]
\[
\frac{dz}{dt}
=
\frac{\partial z}{\partial x}\frac{dx}{dt}
+
\frac{\partial z}{\partial y}\frac{dy}{dt}
\]
Diagram 2: Two-Level Dependency
z
x
y
u
v
w
This represents
\[
z = f(x,y), \quad x = x(u,v), \quad y = y(w)
\]
\[
\frac{\partial z}{\partial u}
=
\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}
\]
\[
\frac{\partial z}{\partial v}
=
\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}
\]
\[
\frac{\partial z}{\partial w}
=
\frac{\partial z}{\partial y}\frac{\partial y}{\partial w}
\]
Diagram 3: Polar Coordinate Chain Rule
z
x
y
r
θ
This corresponds to
\[
x = r\cos\theta, \qquad y = r\sin\theta
\]
\[
\frac{\partial z}{\partial r}
=
\frac{\partial z}{\partial x}\cos\theta
+
\frac{\partial z}{\partial y}\sin\theta
\]
Advanced Examples on Total Derivatives
Solved Examples
Example 1
Find \(dz/dt\) if
\(z = x^2y + y^3\)
where
\(x=t^2,\quad y=t^3\)
Show Solution
\[
\frac{\partial z}{\partial x}=2xy
\]
\[
\frac{\partial z}{\partial y}=x^2+3y^2
\]
\[
\frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=3t^2
\]
Using total derivative formula
\[
\frac{dz}{dt}=(2xy)(2t)+(x^2+3y^2)(3t^2)
\]
Substitute \(x=t^2\), \(y=t^3\)
\[
\frac{dz}{dt}=7t^6+9t^8
\]
Example 2
If
\(z=\ln(x^2+y^2)\)
where
\(x=r\cos\theta,\quad y=r\sin\theta\)
Find \(\partial z/\partial r\).
Show Solution
Since
\[
x^2+y^2=r^2
\]
Thus
\[
z=\ln(r^2)
\]
Differentiate
\[
\frac{\partial z}{\partial r}=\frac{2}{r}
\]
Example 3
Find \(dz/dt\) if
\(z=e^{xy}\)
where
\(x=t^2,\quad y=t\)
Show Solution
\[
\frac{\partial z}{\partial x}=ye^{xy}
\]
\[
\frac{\partial z}{\partial y}=xe^{xy}
\]
\[
\frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=1
\]
\[
\frac{dz}{dt}=ye^{xy}(2t)+xe^{xy}
\]
Substitute \(x=t^2\), \(y=t\).
Example 4.
If \( z = x^2y + y^3 \), where \( x = t^2 \) and \( y = t^3 \), find \( \frac{dz}{dt} \).
Show Solution
\[
z = x^2y + y^3
\]
Partial derivatives:
\[
\frac{\partial z}{\partial x}=2xy
\]
\[
\frac{\partial z}{\partial y}=x^2+3y^2
\]
Since
\[
x=t^2, \quad y=t^3
\]
\[
\frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2
\]
Total derivative:
\[
\frac{dz}{dt}
=\frac{\partial z}{\partial x}\frac{dx}{dt}
+\frac{\partial z}{\partial y}\frac{dy}{dt}
\]
Substitute:
\[
=2xy(2t)+(x^2+3y^2)(3t^2)
\]
Replacing \(x=t^2, y=t^3\):
\[
=4t(t^2t^3)+3t^2(t^4+3t^6)
\]
\[
=4t^6+3t^6+9t^8
\]
\[
\boxed{\frac{dz}{dt}=7t^6+9t^8}
\]
Example 5.
If \( z = e^{xy} \), where \( x = r\cos\theta \) and \( y = r\sin\theta \), find \( \frac{\partial z}{\partial r} \).
Show Solution
\[
z=e^{xy}
\]
Partial derivatives:
\[
\frac{\partial z}{\partial x}=ye^{xy}
\]
\[
\frac{\partial z}{\partial y}=xe^{xy}
\]
Now
\[
\frac{\partial x}{\partial r}=\cos\theta
\]
\[
\frac{\partial y}{\partial r}=\sin\theta
\]
Total derivative:
\[
\frac{\partial z}{\partial r}
=
\frac{\partial z}{\partial x}\frac{\partial x}{\partial r}
+
\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}
\]
Substitute:
\[
=ye^{xy}\cos\theta + xe^{xy}\sin\theta
\]
Factor:
\[
=e^{xy}(y\cos\theta+x\sin\theta)
\]
Replace \(x=r\cos\theta\), \(y=r\sin\theta\)
\[
=e^{r^2\sin\theta\cos\theta}(r\sin\theta\cos\theta+r\cos\theta\sin\theta)
\]
\[
\boxed{\frac{\partial z}{\partial r}=2r\sin\theta\cos\theta \; e^{r^2\sin\theta\cos\theta}}
\]
Example 6.
If \( z = \ln(x^2+y^2) \), where \( x=t^2+1 \) and \( y=2t \), find \( \frac{dz}{dt} \).
Show Solution
\[
z=\ln(x^2+y^2)
\]
Partial derivatives:
\[
\frac{\partial z}{\partial x}=\frac{2x}{x^2+y^2}
\]
\[
\frac{\partial z}{\partial y}=\frac{2y}{x^2+y^2}
\]
Now
\[
\frac{dx}{dt}=2t
\]
\[
\frac{dy}{dt}=2
\]
Total derivative:
\[
\frac{dz}{dt}
=
\frac{2x}{x^2+y^2}(2t)
+
\frac{2y}{x^2+y^2}(2)
\]
\[
=
\frac{4xt+4y}{x^2+y^2}
\]
Substitute \(x=t^2+1, y=2t\):
\[
\boxed{\frac{dz}{dt}=
\frac{4t(t^2+1)+8t}{(t^2+1)^2+4t^2}}
\]
Example 7.
If \( z = x^3y^2 \), where \( x = u+v \) and \( y = uv \), find \( \frac{\partial z}{\partial u} \).
Show Solution
\[
z=x^3y^2
\]
Partial derivatives:
\[
\frac{\partial z}{\partial x}=3x^2y^2
\]
\[
\frac{\partial z}{\partial y}=2x^3y
\]
Now
\[
\frac{\partial x}{\partial u}=1
\]
\[
\frac{\partial y}{\partial u}=v
\]
Total derivative:
\[
\frac{\partial z}{\partial u}
=
3x^2y^2(1)
+
2x^3y(v)
\]
\[
=3x^2y^2+2vx^3y
\]
Substitute \(x=u+v\), \(y=uv\):
\[
\boxed{
\frac{\partial z}{\partial u}
=
3(u+v)^2(u^2v^2)+2v(u+v)^3(uv)
}
\]
Example 8.
If \( z = \sin(xy) \), where \( x=r^2 \), \( y=r^3 \), find \( \frac{dz}{dr} \).
Show Solution
\[
z=\sin(xy)
\]
\[
\frac{\partial z}{\partial x}=y\cos(xy)
\]
\[
\frac{\partial z}{\partial y}=x\cos(xy)
\]
Now
\[
\frac{dx}{dr}=2r
\]
\[
\frac{dy}{dr}=3r^2
\]
Total derivative:
\[
\frac{dz}{dr}
=
y\cos(xy)(2r)+x\cos(xy)(3r^2)
\]
\[
=\cos(xy)(2ry+3r^2x)
\]
Substitute \(x=r^2, y=r^3\):
\[
\boxed{\frac{dz}{dr}
=
\cos(r^5)(2r^4+3r^4)
=
5r^4\cos(r^5)}
\]
Example 9.
If \( z = x^2+y^2 \), where \( x = \cos t \) and \( y = \sin t \), find \( \frac{dz}{dt} \).
Show Solution
\[
z=x^2+y^2
\]
\[
\frac{\partial z}{\partial x}=2x
\]
\[
\frac{\partial z}{\partial y}=2y
\]
Now
\[
\frac{dx}{dt}=-\sin t
\]
\[
\frac{dy}{dt}=\cos t
\]
Total derivative:
\[
\frac{dz}{dt}
=
2x(-\sin t)+2y(\cos t)
\]
Substitute \(x=\cos t, y=\sin t\):
\[
=2\cos t(-\sin t)+2\sin t(\cos t)
\]
\[
\boxed{\frac{dz}{dt}=0}
\]
Example 10.
If \( z = e^{x+y} \), where \( x=t^2 \) and \( y=\ln t \), find \( \frac{dz}{dt} \).
Show Solution
\[
z=e^{x+y}
\]
Partial derivatives:
\[
\frac{\partial z}{\partial x}=e^{x+y}
\]
\[
\frac{\partial z}{\partial y}=e^{x+y}
\]
Now
\[
\frac{dx}{dt}=2t
\]
\[
\frac{dy}{dt}=\frac{1}{t}
\]
Total derivative:
\[
\frac{dz}{dt}
=
e^{x+y}(2t)+e^{x+y}\left(\frac{1}{t}\right)
\]
\[
=e^{x+y}\left(2t+\frac{1}{t}\right)
\]
Substitute \(x=t^2\), \(y=\ln t\):
\[
\boxed{\frac{dz}{dt}=e^{t^2+\ln t}\left(2t+\frac{1}{t}\right)}
\]
