TANSCHE SMTPC12 Differential Calculus Unit 2

Differential Calculus
(TANSCHE Syllabus)
Madurai Kamaraj University
I- B.Sc., Mathematics, I - Semester

ARO Study Circle

2 Partial Differentiation

Syllabus

Partial derivatives – Successive partial derivatives – Function of a function rule – Total differential coefficient – A special case – Implicit Functions.
Chapter 8 Sections 1.1 – 1.5

2.1 Partial Differentiation

So far, we have studied functions of a single variable, such as \(y = f(x)\). However, in real-world problems, we often encounter functions that depend on more than one variable.

For instance, the area of a rectangle depends on both its length and breadth. If we denote the area by \(A\), the length by \(x\), and the breadth by \(y\), then \[A = xy\] Here, \(A\) is a function of two independent variables \(x\) and \(y\). Similarly, the volume of a rectangular box depends on three variables: length, breadth, and height.

2.1.1 What is Partial Differentiation?

When we have a function \(u = f(x,y)\) of two independent variables \(x\) and \(y\), we can differentiate it with respect to one variable while keeping the other variable constant. This process is called partial differentiation.

2.1.1.1 Why do we need Partial Derivatives?

When we change \(x\) while keeping \(y\) fixed, the function \(u\) changes. The rate of this change is measured by the partial derivative of \(u\) with respect to \(x\). Similarly, when we change \(y\) while keeping \(x\) fixed, we get the partial derivative of \(u\) with respect to \(y\).

2.1.2 Definition of Partial Derivatives

2.1.2.1 Partial Derivative with respect to \(x\)

Let \(u = f(x,y)\) be a function of two variables. The partial derivative of \(u\) with respect to \(x\), denoted by \(\dfrac{\partial u}{\partial x}\) or \(f_x\), is defined as \[\dfrac{\partial u}{\partial x} = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x, y) - f(x,y)}{\Delta x}\]

This means:

• We keep \(y\) constant
• We allow \(x\) to change by a small amount \(\Delta x\)
• We find the corresponding change in \(u\)
• We divide by \(\Delta x\) and take the limit as \(\Delta x \to 0\)

2.1.2.2 Partial Derivative with respect to \(y\)

Similarly, the partial derivative of \(u\) with respect to \(y\), denoted by \(\dfrac{\partial u}{\partial y}\) or \(f_y\), is defined as \[\dfrac{\partial u}{\partial y} = \lim_{\Delta y \to 0} \dfrac{f(x, y+\Delta y) - f(x,y)}{\Delta y}\]

Here:

• We keep \(x\) constant
• We allow \(y\) to change by a small amount \(\Delta y\)
• We find the corresponding change in \(u\)
• We divide by \(\Delta y\) and take the limit as \(\Delta y \to 0\)

2.1.2.3 Notation

The partial derivative \(\dfrac{\partial u}{\partial x}\) can also be written as \[\dfrac{\partial}{\partial x} f(x,y) \quad \text{or} \quad \dfrac{\partial f}{\partial x} \quad \text{or} \quad f_x\]

Similarly, \(\dfrac{\partial u}{\partial y}\) can be written as \[\dfrac{\partial}{\partial y} f(x,y) \quad \text{or} \quad \dfrac{\partial f}{\partial y} \quad \text{or} \quad f_y\]

Important Note: The symbol \(\partial\) (called "del" or "partial") is used to indicate partial differentiation, distinguishing it from ordinary differentiation denoted by \(d\).

2.1.3 Successive Partial Derivatives

Just like ordinary derivatives, partial derivatives can be differentiated again. This gives us higher-order partial derivatives.

2.1.3.1 Pure Second Order Partial Derivatives

If we differentiate \(\dfrac{\partial u}{\partial x}\) again with respect to \(x\) (keeping \(y\) constant), we get \[\dfrac{\partial^2 u}{\partial x^2} = \dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}\right)\]

Similarly, if we differentiate \(\dfrac{\partial u}{\partial y}\) again with respect to \(y\) (keeping \(x\) constant), we get \[\dfrac{\partial^2 u}{\partial y^2} = \dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial y}\right)\]

2.1.3.2 Mixed Second Order Partial Derivatives

If we first differentiate \(u\) with respect to \(x\) and then differentiate the result with respect to \(y\), we get \[\dfrac{\partial^2 u}{\partial y \partial x} = \dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial x}\right)\]

If we first differentiate \(u\) with respect to \(y\) and then differentiate the result with respect to \(x\), we get \[\dfrac{\partial^2 u}{\partial x \partial y} = \dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial y}\right)\]

2.1.3.3 Important Result

For most functions that we encounter in mathematics and physics, the mixed partial derivatives are equal: \[\boxed{\dfrac{\partial^2 u}{\partial y \partial x} = \dfrac{\partial^2 u}{\partial x \partial y}}\]

This means the order of differentiation does not matter. This result is known as Clairaut’s Theorem or the Equality of Mixed Partials.

2.2 Function of Function Rule (or) The Chain Rule for Partial Derivatives

When we have a composite function, we need the chain rule for partial differentiation.

2.2.0.1 Statement of the Rule

Let \(z\) be a function of \(u\), where \(u\) is a function of two independent variables \(x\) and \(y\). Then, \[\dfrac{\partial z}{\partial x} = \dfrac{dz}{du} \dfrac{\partial u}{\partial x}\] and \[\dfrac{\partial z}{\partial y} = \dfrac{dz}{du} \dfrac{\partial u}{\partial y}\]

2.2.0.2 Understanding the Rule

• \(z\) depends on \(u\) only: \(z = F(u)\)
• \(u\) depends on \(x\) and \(y\): \(u = f(x,y)\)
• To find \(\dfrac{\partial z}{\partial x}\), we first find \(\dfrac{dz}{du}\) (ordinary derivative) and multiply by \(\dfrac{\partial u}{\partial x}\) (partial derivative)
• To find \(\dfrac{\partial z}{\partial y}\), we multiply \(\dfrac{dz}{du}\) by \(\dfrac{\partial u}{\partial y}\)

Note on Notation: - We use \(d\) for \(\dfrac{dz}{du}\) because \(z\) depends on only one variable \(u\) - We use \(\partial\) for \(\dfrac{\partial u}{\partial x}\) because \(u\) depends on two variables \(x\) and \(y\)

2.2.1 Illustrative Examples

Example 1.1. Find the partial derivatives of \(u = \sin(ax + by + cz)\).

Solution . Here \(u\) is a function of three variables \(x, y, z\).

To find \(\dfrac{\partial u}{\partial x}\), we treat \(y\) and \(z\) as constants: \[\begin{align*} \dfrac{\partial u}{\partial x} &= a \cos(ax + by + cz) \end{align*}\]

Similarly, \[\begin{align*} \dfrac{\partial u}{\partial y} &= b \cos(ax + by + cz) \\ \dfrac{\partial u}{\partial z} &= c \cos(ax + by + cz) \end{align*}\]

Example 1.2. If \(u = \dfrac{xy}{x+y}\), show that \[x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = u\]

Solution . First, we find \(\dfrac{\partial u}{\partial x}\). Using the quotient rule (treating \(y\) as constant): \[\begin{align*} \dfrac{\partial u}{\partial x} &= \dfrac{(x+y)\dfrac{\partial}{\partial x}(xy) - xy\dfrac{\partial}{\partial x}(x+y)}{(x+y)^2} \\ &= \dfrac{(x+y)y - xy(1)}{(x+y)^2} \\ &= \dfrac{xy + y^2 - xy}{(x+y)^2} \\ &= \dfrac{y^2}{(x+y)^2} \end{align*}\]

Similarly, \[\begin{align*} \dfrac{\partial u}{\partial y} &= \dfrac{(x+y)\dfrac{\partial}{\partial y}(xy) - xy\dfrac{\partial}{\partial y}(x+y)}{(x+y)^2} \\ &= \dfrac{(x+y)x - xy(1)}{(x+y)^2} \\ &= \dfrac{x^2 + xy - xy}{(x+y)^2} \\ &= \dfrac{x^2}{(x+y)^2} \end{align*}\]

Now, \[\begin{align*} x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} &= x\dfrac{y^2}{(x+y)^2} + y\dfrac{x^2}{(x+y)^2} \\ &= \dfrac{xy^2 + x^2y}{(x+y)^2} \\ &= \dfrac{xy(x+y)}{(x+y)^2} \\ &= \dfrac{xy}{x+y} = u \end{align*}\]

\[\boxed{x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = u}\]

Example 1.3. If \(u = \tan^{-1}\left(\dfrac{x^3+y^3}{x-y}\right)\), prove that \[x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = \sin 2u\]

Solution. Let \(\tan u = \dfrac{x^3+y^3}{x-y}\). Let \(v = \tan u\).

Differentiating \(\tan u = \dfrac{x^3+y^3}{x-y}\) with respect to \(x\) (treating \(y\) as constant): \[\begin{align*} \sec^2 u \dfrac{\partial u}{\partial x} &= \dfrac{(x-y)3x^2 - (x^3+y^3)(1)}{(x-y)^2} \\ &= \dfrac{3x^3 - 3x^2y - x^3 - y^3}{(x-y)^2} \\ &= \dfrac{2x^3 - 3x^2y - y^3}{(x-y)^2} \end{align*}\]

Differentiating with respect to \(y\) (treating \(x\) as constant): \[\begin{align*} \sec^2 u \dfrac{\partial u}{\partial y} &= \dfrac{(x-y)3y^2 - (x^3+y^3)(-1)}{(x-y)^2} \\ &= \dfrac{3xy^2 - 3y^3 + x^3 + y^3}{(x-y)^2} \\ &= \dfrac{x^3 + 3xy^2 - 2y^3}{(x-y)^2} \end{align*}\]

Now, \[\begin{align*} \sec^2 u \left(x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y}\right) &= x\dfrac{2x^3 - 3x^2y - y^3}{(x-y)^2} + y\dfrac{x^3 + 3xy^2 - 2y^3}{(x-y)^2} \\ &= \dfrac{2x^4 - 3x^3y - xy^3 + x^3y + 3xy^3 - 2y^4}{(x-y)^2} \\ &= \dfrac{2x^4 - 2x^3y + 2xy^3 - 2y^4}{(x-y)^2} \\ &= \dfrac{2(x^4 - x^3y + xy^3 - y^4)}{(x-y)^2} \\ &= \dfrac{2[(x^3)(x-y) + y^3(x-y)]}{(x-y)^2} \\ &= \dfrac{2(x^3+y^3)(x-y)}{(x-y)^2} \\ &= \dfrac{2(x^3+y^3)}{x-y} \\ &= 2\tan u \end{align*}\]

Therefore, \[\begin{align*} x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} &= \dfrac{2\tan u}{\sec^2 u} \\ &= 2\tan u \cos^2 u \\ &= 2\sin u \cos u \\ &= \sin 2u \end{align*}\]

\[\boxed{x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = \sin 2u}\]

Example 1.4. If \(V = (x^2 + y^2 + z^2)^{-\dfrac{1}{2}}\), show that \[\dfrac{\partial^2 V}{\partial x^2} + \dfrac{\partial^2 V}{\partial y^2} + \dfrac{\partial^2 V}{\partial z^2} = 0\]

Solution. Given, \[V = (x^2 + y^2 + z^2)^{-\dfrac{1}{2}}\]

Differentiating \(V\) with respect to \(x\) alone (treating \(y\) and \(z\) as constants): \[\begin{align*} \dfrac{\partial V}{\partial x} &= -\dfrac{1}{2}(x^2 + y^2 + z^2)^{-\dfrac{3}{2}} \cdot 2x \\ &= -x(x^2 + y^2 + z^2)^{-\dfrac{3}{2}} \end{align*}\]

Differentiating once again with respect to \(x\) alone: \[\begin{align*} \dfrac{\partial^2 V}{\partial x^2} &= \dfrac{\partial}{\partial x}\left[-x(x^2 + y^2 + z^2)^{-\dfrac{3}{2}}\right] \\ &= -(x^2 + y^2 + z^2)^{-\dfrac{3}{2}} + (-x)\left(-\dfrac{3}{2}\right)(x^2 + y^2 + z^2)^{-\dfrac{5}{2}} \cdot 2x \\ &= -(x^2 + y^2 + z^2)^{-\dfrac{3}{2}} + 3x^2(x^2 + y^2 + z^2)^{-\dfrac{5}{2}} \\ &= \dfrac{- (x^2 + y^2 + z^2) + 3x^2}{(x^2 + y^2 + z^2)^{\dfrac{5}{2}}} \\ &= \dfrac{2x^2 - y^2 - z^2}{(x^2 + y^2 + z^2)^{\dfrac{5}{2}}} \end{align*}\]

Similarly, \[\begin{align*} \dfrac{\partial^2 V}{\partial y^2} &= \dfrac{2y^2 - z^2 - x^2}{(x^2 + y^2 + z^2)^{\dfrac{5}{2}}} \\ \dfrac{\partial^2 V}{\partial z^2} &= \dfrac{2z^2 - x^2 - y^2}{(x^2 + y^2 + z^2)^{\dfrac{5}{2}}} \end{align*}\]

Therefore, \[\begin{align*} \dfrac{\partial^2 V}{\partial x^2} + \dfrac{\partial^2 V}{\partial y^2} + \dfrac{\partial^2 V}{\partial z^2} &= \dfrac{(2x^2 - y^2 - z^2) + (2y^2 - z^2 - x^2) + (2z^2 - x^2 - y^2)}{(x^2 + y^2 + z^2)^{\dfrac{5}{2}}} \\ &= \dfrac{2x^2 + 2y^2 + 2z^2 - x^2 - y^2 - z^2 - x^2 - y^2 - z^2}{(x^2 + y^2 + z^2)^{\dfrac{5}{2}}} \\ &= \dfrac{0}{(x^2 + y^2 + z^2)^{\dfrac{5}{2}}} \\ &= 0 \end{align*}\]

\[\boxed{\dfrac{\partial^2 V}{\partial x^2} + \dfrac{\partial^2 V}{\partial y^2} + \dfrac{\partial^2 V}{\partial z^2} = 0}\]

(V is said to satisfy Laplace’s equation.)

Example 1.5. Illustrate the theorem that \(\dfrac{\partial^2 u}{\partial x \partial y} = \dfrac{\partial^2 u}{\partial y \partial x}\) where \[u = \log \dfrac{x^2 + y^2}{xy}\]

Solution . Given, \[u = \log \dfrac{x^2 + y^2}{xy}\]

Using properties of logarithms: \[u = \log(x^2 + y^2) - \log x - \log y\]

Differentiating \(u\) with respect to \(x\) (treating \(y\) as constant): \[\begin{align*} \dfrac{\partial u}{\partial x} &= \dfrac{2x}{x^2 + y^2} - \dfrac{1}{x} \end{align*}\]

Now, differentiate \(\dfrac{\partial u}{\partial x}\) with respect to \(y\) (treating \(x\) as constant): \[\begin{align*} \dfrac{\partial^2 u}{\partial y \partial x} &= \dfrac{\partial}{\partial y}\left(\dfrac{2x}{x^2 + y^2} - \dfrac{1}{x}\right) \\ &= \dfrac{\partial}{\partial y}\left(\dfrac{2x}{x^2 + y^2}\right) - \dfrac{\partial}{\partial y}\left(\dfrac{1}{x}\right) \\ &= 2x \cdot \dfrac{\partial}{\partial y}(x^2 + y^2)^{-1} - 0 \\ &= 2x \cdot (-1)(x^2 + y^2)^{-2} \cdot 2y \\ &= -\dfrac{4xy}{(x^2 + y^2)^2} \end{align*}\]

Differentiating \(u\) with respect to \(y\) (treating \(x\) as constant): \[\begin{align*} \dfrac{\partial u}{\partial y} &= \dfrac{2y}{x^2 + y^2} - \dfrac{1}{y} \end{align*}\]

Now, differentiate \(\dfrac{\partial u}{\partial y}\) with respect to \(x\) (treating \(y\) as constant): \[\begin{align*} \dfrac{\partial^2 u}{\partial x \partial y} &= \dfrac{\partial}{\partial x}\left(\dfrac{2y}{x^2 + y^2} - \dfrac{1}{y}\right) \\ &= \dfrac{\partial}{\partial x}\left(\dfrac{2y}{x^2 + y^2}\right) - \dfrac{\partial}{\partial x}\left(\dfrac{1}{y}\right) \\ &= 2y \cdot \dfrac{\partial}{\partial x}(x^2 + y^2)^{-1} - 0 \\ &= 2y \cdot (-1)(x^2 + y^2)^{-2} \cdot 2x \\ &= -\dfrac{4xy}{(x^2 + y^2)^2} \end{align*}\]

Therefore, \[\dfrac{\partial^2 u}{\partial y \partial x} = -\dfrac{4xy}{(x^2 + y^2)^2} = \dfrac{\partial^2 u}{\partial x \partial y}\]

\[\boxed{\dfrac{\partial^2 u}{\partial y \partial x} = \dfrac{\partial^2 u}{\partial x \partial y}}\]

Hence, the theorem \(\dfrac{\partial^2 u}{\partial x \partial y} = \dfrac{\partial^2 u}{\partial y \partial x}\) is verified.

2.3 Total Differential Coefficient

So far, we have considered partial derivatives where we differentiate with respect to one variable while keeping the other variables constant. However, sometimes both variables change simultaneously. In such cases, we need the total differential coefficient.

Let \(u = f(x,y)\) be a continuous function of two independent variables \(x\) and \(y\). If \(x\) and \(y\) receive small increments \(\Delta x\) and \(\Delta y\), then \(u\) will receive a corresponding small increment \(\Delta u\).

The total increment of \(u\) is given by \[\Delta u = f(x + \Delta x, y + \Delta y) - f(x,y)\]

2.3.1 Derivation of Total Differential

We can write \(\Delta u\) as \[\Delta u = f(x + \Delta x, y + \Delta y) - f(x,y + \Delta y) + f(x,y + \Delta y) - f(x,y)\]

Applying the Mean Value Theorem to each of the two differences:

For the first difference: \[f(x + \Delta x, y + \Delta y) - f(x,y + \Delta y) = f_x(x + \theta_1 \Delta x, y + \Delta y) \Delta x\]

For the second difference: \[f(x,y + \Delta y) - f(x,y) = f_y(x,y + \theta_2 \Delta y) \Delta y\]

where \(f_x\) and \(f_y\) denote the partial derivatives with respect to \(x\) and \(y\) respectively, and \(\theta_1\), \(\theta_2\) are positive fractions between 0 and 1.

Therefore, \[\Delta u = f_x(x + \theta_1 \Delta x, y + \Delta y) \Delta x + f_y(x,y + \theta_2 \Delta y) \Delta y\]

2.3.2 Total Differential Coefficient with respect to a Parameter

Now, suppose \(x\) and \(y\) are continuous functions of some other variable \(t\). Let \(\Delta x\), \(\Delta y\), and \(\Delta u\) be the increments in \(x\), \(y\), and \(u\) due to an increment \(\Delta t\) in \(t\).

Dividing the previous equation by \(\Delta t\): \[\dfrac{\Delta u}{\Delta t} = f_x(x + \theta_1 \Delta x, y + \Delta y) \dfrac{\Delta x}{\Delta t} + f_y(x,y + \theta_2 \Delta y) \dfrac{\Delta y}{\Delta t}\]

Now, let \(\Delta t \to 0\). Then \(\Delta x \to 0\) and \(\Delta y \to 0\), and we get: \[\dfrac{du}{dt} = f_x(x,y) \dfrac{dx}{dt} + f_y(x,y) \dfrac{dy}{dt}\]

\[\boxed{\dfrac{du}{dt} = \dfrac{\partial u}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial u}{\partial y} \dfrac{dy}{dt}}\]

In differential form, this can be written as: \[\boxed{du = \dfrac{\partial u}{\partial x} dx + \dfrac{\partial u}{\partial y} dy}\]

This quantity \(du\) is called the total differential of \(u\).

2.3.3 Extension to More Variables

If \(u = f(x_1, x_2, \ldots, x_n)\) and \(x_1, x_2, \ldots, x_n\) are all functions of \(t\), then \[\dfrac{du}{dt} = \dfrac{\partial u}{\partial x_1} \dfrac{dx_1}{dt} + \dfrac{\partial u}{\partial x_2} \dfrac{dx_2}{dt} + \cdots + \dfrac{\partial u}{\partial x_n} \dfrac{dx_n}{dt}\]

In differential form, \[du = \dfrac{\partial u}{\partial x_1} dx_1 + \dfrac{\partial u}{\partial x_2} dx_2 + \cdots + \dfrac{\partial u}{\partial x_n} dx_n\]

2.4 A Special Case

If \(u = f(x,y)\) where \(x\) and \(y\) are functions of \(t\), we get: \[\dfrac{du}{dt} = \dfrac{\partial u}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial u}{\partial y} \dfrac{dy}{dt}\]

Now, if we take \(t = x\), then \(u\) becomes a function of \(x\) and \(y\), where \(y\) is a function of \(x\). Since \(\dfrac{dx}{dx} = 1\), the relation becomes: \[\boxed{\dfrac{du}{dx} = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} \dfrac{dy}{dx}}\]

Important Distinction: The quantities \(\dfrac{du}{dx}\) and \(\dfrac{\partial u}{\partial x}\) are quite distinct. - \(\dfrac{du}{dx}\) is the total derivative - it takes into account the fact that \(y\) also changes with \(x\) - \(\dfrac{\partial u}{\partial x}\) is the partial derivative - it assumes \(y\) is constant

Example: Let \(u = x^2 + 2xy + y^2\) and let \(y\) be a function of \(x\). Then, \[\dfrac{du}{dx} = 2x + 2y + (2x + 2y)\dfrac{dy}{dx}\]

The value of \(\dfrac{dy}{dx}\) will depend on the relation between \(x\) and \(y\).

2.5 Implicit Functions

Sometimes the relation between \(x\) and \(y\) is given in implicit form: \[f(x,y) = c\] where \(c\) is a constant.

To find \(\dfrac{dy}{dx}\), we use the total derivative. Since \(f(x,y) = c\) is constant, its total differential is zero: \[\dfrac{\partial f}{\partial x} dx + \dfrac{\partial f}{\partial y} dy = 0\]

Dividing throughout by \(dx\): \[\dfrac{\partial f}{\partial x} + \dfrac{\partial f}{\partial y} \dfrac{dy}{dx} = 0\]

Therefore, \[\boxed{\dfrac{dy}{dx} = -\dfrac{\dfrac{\partial f}{\partial x}}{\dfrac{\partial f}{\partial y}}}\]

provided \(\dfrac{\partial f}{\partial y} \neq 0\).

Illustrative Examples

Example 1.6. Find \(\dfrac{du}{dt}\) where \(u = x^2 + y^2 + z^2\), \(x = e^t\), \(y = e^t \sin t\), and \(z = e^t \cos t\).

Solution . Given \(u = x^2 + y^2 + z^2\).

First, find the partial derivatives of \(u\): \[\begin{align*} \dfrac{\partial u}{\partial x} = 2x, \qquad \dfrac{\partial u}{\partial y} = 2y, \qquad \dfrac{\partial u}{\partial z} = 2z \end{align*}\]

Now, find the derivatives of \(x, y, z\) with respect to \(t\): \[\begin{align*} \dfrac{dx}{dt} = e^t , \qquad \dfrac{dy}{dt} = e^t \sin t + e^t \cos t = e^t(\sin t + \cos t) , \qquad \dfrac{dz}{dt} = e^t \cos t - e^t \sin t = e^t(\cos t - \sin t) \end{align*}\]

Using the total derivative formula: \[\dfrac{du}{dt} = \dfrac{\partial u}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial u}{\partial y} \dfrac{dy}{dt} + \dfrac{\partial u}{\partial z} \dfrac{dz}{dt}\]

Substituting the values: \[\begin{align*} \dfrac{du}{dt} &= 2x(e^t) + 2y(e^t \sin t + e^t \cos t) + 2z(e^t \cos t - e^t \sin t) \\ &= 2e^t(x + y \sin t + y \cos t + z \cos t - z \sin t) \end{align*}\]

Now, substitute \(x = e^t\), \(y = e^t \sin t\), \(z = e^t \cos t\): \[\begin{align*} \dfrac{du}{dt} &= 2e^t(e^t + e^t \sin^2 t + e^t \sin t \cos t + e^t \cos^2 t - e^t \cos t \sin t) \\ &= 2e^t(e^t + e^t \sin^2 t + e^t \cos^2 t) \\ &= 2e^t(e^t + e^t(\sin^2 t + \cos^2 t)) \\ &= 2e^t(e^t + e^t) \\ &= 2e^t(2e^t) \\ &= 4e^{2t} \end{align*}\]

\[\boxed{\dfrac{du}{dt} = 4e^{2t}}\]

Example 1.7. Find \(\dfrac{du}{dx}\) when \(u = x^2 + y^2\) where \(y = \dfrac{1-x}{x}\).

Solution . Given \(u = x^2 + y^2\).

Here \(u\) is a function of \(x\) and \(y\), where \(y\) is a function of \(x\). So we use the special case formula: \[\dfrac{du}{dx} = \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} \dfrac{dy}{dx}\]

First, find the partial derivatives of \(u\): \[\begin{align*} \dfrac{\partial u}{\partial x} = 2x , \qquad \dfrac{\partial u}{\partial y} = 2y \end{align*}\]

Now, find \(\dfrac{dy}{dx}\): \[\begin{align*} y &= \dfrac{1-x}{x} = \dfrac{1}{x} - 1 \\ \dfrac{dy}{dx} &= -\dfrac{1}{x^2} \end{align*}\]

Substituting in the formula: \[\begin{align*} \dfrac{du}{dx} &= 2x + 2y\left(-\dfrac{1}{x^2}\right) \\ &= 2x - \dfrac{2y}{x^2} \end{align*}\]

Substitute \(y = \dfrac{1-x}{x}\): \[\begin{align*} \dfrac{du}{dx} &= 2x - \dfrac{2(1-x)}{x^2 \cdot x} \\ &= 2x - \dfrac{2(1-x)}{x^3} \\ &= \dfrac{2x^4 - 2(1-x)}{x^3} \\ &= \dfrac{2x^4 - 2 + 2x}{x^3} \\ &= \dfrac{2(x^4 + x - 1)}{x^3} \end{align*}\]

\[\boxed{\dfrac{du}{dx} = \dfrac{2(x^4 + x - 1)}{x^3}}\]

Example 1.8. If \(x^3 + y^3 - 3axy = 0\), find \(\dfrac{dy}{dx}\).

Solution . Given, \[x^3 + y^3 - 3axy = 0\]

Let \(f(x,y) = x^3 + y^3 - 3axy = 0\).

This is an implicit function. To find \(\dfrac{dy}{dx}\), we use the formula: \[\dfrac{dy}{dx} = -\dfrac{\dfrac{\partial f}{\partial x}}{\dfrac{\partial f}{\partial y}}\]

First, find the partial derivatives of \(f\): \[\begin{align*} \dfrac{\partial f}{\partial x} &= 3x^2 - 3ay \\ \dfrac{\partial f}{\partial y} &= 3y^2 - 3ax \end{align*}\]

Now, substitute in the formula: \[\begin{align*} \dfrac{dy}{dx} &= -\dfrac{3x^2 - 3ay}{3y^2 - 3ax} \\ &= -\dfrac{3(x^2 - ay)}{3(y^2 - ax)} \\ &= -\dfrac{x^2 - ay}{y^2 - ax} \end{align*}\]

\[\boxed{\dfrac{dy}{dx} = -\dfrac{x^2 - ay}{y^2 - ax}}\]