Problem 28 - Calculus with Concepts in Calculus

Question

Find \(d y / d x\) by implicit differentiation. $$ e^{x / y}+\ln y / x+15=0 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1 - \frac{e^{x/y}}{y}}{\frac{x}{y} - \frac{x \cdot e^{x/y}}{y^2}} \)

Step 1: Differentiate Each Term

First, differentiate each term of the equation with respect to \(x\), remembering that \(y\) is also a function of \(x\) (i.e., \(y = y(x)\)).For the term \(e^{x/y}\), use the chain rule:\[ \frac{d}{dx} \left( e^{x/y} \right) = e^{x/y} \cdot \frac{d}{dx} \left( \frac{x}{y} \right) \]Differentiating \(\frac{x}{y}\), we use the quotient rule:\[ \frac{d}{dx} \left( \frac{x}{y} \right) = \frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} = \frac{y - x \cdot \frac{dy}{dx}}{y^2} \]Therefore,\[ \frac{d}{dx} \left( e^{x/y} \right) = e^{x/y} \cdot \frac{y - x \cdot \frac{dy}{dx}}{y^2} \]

Step 2: Differentiate the Natural Logarithm Term

For the term \( \ln \left( \frac{y}{x} \right) \), differentiate using the chain rule and quotient rule:\[ \frac{d}{dx} \left( \ln \left( \frac{y}{x} \right) \right) = \frac{1}{y/x} \cdot \frac{d}{dx} \left( \frac{y}{x} \right) \]Differentiating \(\frac{y}{x}\), we apply the quotient rule:\[ \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{x \cdot \frac{dy}{dx} - y \cdot 1}{x^2} = \frac{x \cdot \frac{dy}{dx} - y}{x^2} \]So, the derivative of \(\ln(y/x)\) is:\[ \frac{x}{y} \cdot \left( \frac{x \cdot \frac{dy}{dx} - y}{x^2} \right) = \frac{1}{y} \cdot \left( x \cdot \frac{dy}{dx} - y \right) \]

Step 3: Differentiate the Constant

The derivative of the constant \(15\) is simply \(0\).

Step 4: Set Up the Equation

Combine all the differentiated components into a single equation:\[ e^{x/y} \cdot \frac{y - x \cdot \frac{dy}{dx}}{y^2} + \frac{1}{y} \cdot (x \cdot \frac{dy}{dx} - y) + 0 = 0 \]

Step 5: Solve for \( \frac{dy}{dx} \)

Combine like terms and simplify the equation to solve for \( \frac{dy}{dx} \):1. Expand the equation: \[ e^{x/y} \cdot \frac{y}{y^2} - e^{x/y} \cdot \frac{x \cdot \frac{dy}{dx}}{y^2} + \frac{x}{y} \cdot \frac{dy}{dx} - 1 = 0 \]2. This simplifies to: \[ \frac{e^{x/y}}{y} - \frac{x \cdot e^{x/y}}{y^2} \cdot \frac{dy}{dx} + \frac{x}{y} \cdot \frac{dy}{dx} - 1 = 0 \]3. Isolate terms containing \(\frac{dy}{dx}\): \[ \left( \frac{x}{y} - \frac{x \cdot e^{x/y}}{y^2} \right) \cdot \frac{dy}{dx} = 1 - \frac{e^{x/y}}{y} \]4. So, we solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{1 - \frac{e^{x/y}}{y}}{\frac{x}{y} - \frac{x \cdot e^{x/y}}{y^2}} \]

Step 6: Simplify the Solution if Necessary

Once you have \( \frac{dy}{dx} \), you can simplify further if needed or leave it as is based on the form presented in the previous step.

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental tool in calculus that allows us to differentiate composite functions. When a function is composed of another function, it's not straightforward to differentiate without appealing to the chain rule. This rule helps basically to "unpack" these layers.
To understand it fully, consider a function within a function: if you have a situation with an outer function denoted by \(f\) and an inner function described by \(g(x)\), the overall function can be expressed as \(f(g(x))\). Here’s how it operates:

• Diffe rentiate the outer function \(f\) while keeping the inner function \(g(x)\) constant.
• Then, multiply it by the derivative of the inner function \(g(x)\).

For the expression \(e^{x/y}\), we apply the chain rule by treating \(x/y\) as the inner function. Therefore, we differentiate \(e^{u}\) as \(e^{u}\), then multiply this by the derivative of \(x/y\).
This process allows us to effectively break down complex nested functions into simpler parts, manage and solve them incrementally.

Quotient Rule in Calculus

The quotient rule is a very useful differentiation technique when dealing with rational functions — that is, functions that are expressed as a quotient of two other functions. When you have a function of the form \(\frac{u(x)}{v(x)}\), applying the quotient rule can simplify finding the derivative, rather than using substitution or other methods.
The quotient rule states that the derivative of \(\frac{u}{v}\) is given by: \[ (\frac{u}{v})' = \frac{v \cdot u' - u \cdot v'}{v^2} \] Here's how it works step-by-step:

• Diffe rentiate the numerator \(u\) to get \(u'\), while keeping the denominator \(v\) constant.
• Diffe rentiate the denominator \(v\) to get \(v'\), while keeping the numerator \(u\) constant.
• Now, apply the formula to find the overall derivative.

In the given exercise, this rule is crucial in differentiating terms like \(\frac{x}{y}\) and \(\frac{y}{x}\). By understanding the quotient rule, the calculation of derivatives involving division becomes systematic and straightforward.

Differentiation Techniques

Differentiation is one of the core processes in calculus, and it calls for certain techniques to handle different types of functions. As in our exercise, implicit differentiation is used when the y-variable is considered to be a function of x, and these two variables are intermixed in an equation.
Here are some essential differentiation techniques that can help solve complex problems:

• Basic differentiation: Know the derivatives of basic functions such as power, exponential, and logarithmic components.
• Chain Rule: Used for composite functions, as previously discussed.
• Product Rule: Available for functions that are multiplicatively combined.
• Quotient Rule: Applicable for functions divided by each other, very useful in rational functions.

Considering these techniques also includes simplifying expressions and sometimes requiring rewriting of functions. In implicit differentiation, like what's done in the original problem, you sometimes solve an equation involving both y and dy/dx.
Combining these techniques can help differentiate a wide range of complex functions and make seemingly tough calculus problems manageable.