Total Derivatives

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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
\(x=t^2,\quad y=t^3\)
\[ \frac{\partial z}{\partial x}=2xy \] \[ \frac{\partial z}{\partial y}=x^2+3y^2 \] \[ \frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=3t^2 \] 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 \(\partial z/\partial r\).

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
\(x=t^2,\quad y=t\)
\[ \frac{\partial z}{\partial x}=ye^{xy} \] \[ \frac{\partial z}{\partial y}=xe^{xy} \] \[ \frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=1 \] \[ \frac{dz}{dt}=ye^{xy}(2t)+xe^{xy} \] Substitute \(x=t^2\), \(y=t\).
Example 4. If \( z = x^2y + y^3 \), where \( x = t^2 \) and \( y = t^3 \), find \( \frac{dz}{dt} \).

\[ z = x^2y + y^3 \] 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 \( \frac{\partial z}{\partial r} \).

\[ z=e^{xy} \] 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 \( \frac{dz}{dt} \).

\[ z=\ln(x^2+y^2) \] 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 \( \frac{\partial z}{\partial u} \).

\[ z=x^3y^2 \] 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 \( \frac{dz}{dr} \).

\[ z=\sin(xy) \] \[ \frac{\partial z}{\partial x}=y\cos(xy) \] \[ \frac{\partial z}{\partial y}=x\cos(xy) \] 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 \( \frac{dz}{dt} \).

\[ z=x^2+y^2 \] \[ \frac{\partial z}{\partial x}=2x \] \[ \frac{\partial z}{\partial y}=2y \] 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 \( \frac{dz}{dt} \).

\[ z=e^{x+y} \] 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 \(x=t^2\), \(y=\ln t\): \[ \boxed{\frac{dz}{dt}=e^{t^2+\ln t}\left(2t+\frac{1}{t}\right)} \]

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