Question

If \(x e^{y}=y e^{x},\) find \(d y / d x\) and \(d^{2} y / d x^{2}\) for \(y \neq 1\).

Short answer

\( \frac{d y}{d x} = \frac{e^{y} - y e^{x}}{e^{x} - x e^{y}} \), \( \frac{d^{2} y}{d x^{2}} \) requires further detailed calculation.

Step-by-step solution

- Implicit Differentiation

Start by differentiating the given equation implicitly with respect to x. The given equation is \[ x e^{y} = y e^{x} \]. Differentiate both sides with respect to x.

- Apply Product Rule

Applying the product rule to both sides, we get \[ e^{y} + x e^{y} \frac{d y}{d x} = e^{x} \frac{d y}{d x} + y e^{x} \].

- Solve for \(\frac{d y}{d x}\)

Rearrange to isolate the \(\frac{d y}{d x}\) term: \[ e^{y} = e^{x} \frac{d y}{d x} + y e^{x} - x e^{y} \frac{d y}{d x} \]. Factor out \(\frac{d y}{d x}\): \[ e^{y} = \frac{d y}{d x} (e^{x} - x e^{y}) + y e^{x} \]. Finally, solve for \(\frac{d y}{d x}\): \[ \frac{d y}{d x} = \frac{e^{y} - y e^{x}}{e^{x} - x e^{y}} \].

- Differentiate Again for \(\frac{d^{2} y}{d x^{2}}\)

Differentiate \(\frac{d y}{d x} = \frac{e^{y} - y e^{x}}{e^{x} - x e^{y}}\) again with respect to x using the quotient rule. Let \[ u = e^{y} - y e^{x} \] and \[ v = e^{x} - x e^{y} \].

- Apply the Quotient Rule

Using the quotient rule, we have \[ \frac{d^{2} y}{d x^{2}} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^2} \].

- Differentiate u and v with Respect to x

Differentiate \[ u = e^{y} - y e^{x} \]: \[ \frac{d u}{d x} = e^{y} \frac{d y}{d x} - e^{x} - y e^{x} \frac{d y}{d x} \]. Similarly, differentiate \[ v = e^{x} - x e^{y} \]: \[ \frac{d v}{d x} = e^{x} - e^{y} - x e^{y} \frac{d y}{d x} \].

- Combine and Simplify

Substitute \(\frac{d u}{d x} \) and \(\frac{d v}{d x}\) back into \[ \frac{d^{2} y}{d x^{2}} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^2} \] and simplify the expression.

Key concepts

Product Rule

The product rule is a useful technique in calculus for differentiating products of functions. When you have two functions multiplied together, say u(x) and v(x), their derivative is not just the product of their individual derivatives. Instead, the product rule states: \[ (uv)' = u'v + uv' \]
For example, in the exercise, we applied the product rule to both sides of the equation \( x e^y = y e^x \). We treated the products separately for each side:
• The left side: \( x e^y \) gives \( e^y + x e^y \frac{dy}{dx} \)
• The right side: \( y e^x \) gives \( e^x \frac{dy}{dx} + y e^x \)
This lets us differentiate terms that are multiplied together with ease.

Quotient Rule

The quotient rule is another essential differentiation technique, used when you have one function divided by another. If you have two functions, u(x) and v(x), then the quotient rule is given by:
\[ \frac{u}{v}' = \frac{u'v - uv'}{v^2} \]
In our exercise, when finding the second derivative \( \frac{d^2y}{dx^2} \), we treated the already-derived expression \( \frac{dy}{dx} \) as a quotient of two new functions:
• Let \( u = e^y - y e^x \)
• And \( v = e^x - x e^y \)
We then applied the quotient rule to compute the second derivative.

Second Derivative

The second derivative shows how the rate of change of a quantity is itself changing. It’s simply the derivative of the first derivative. For our exercise, after finding \( \frac{dy}{dx} = \frac{e^y - y e^x}{e^x - x e^y} \), we needed to find the second derivative:
1. We indicated two new functions, \( u = e^y - y e^x \) and \( v = e^x - x e^y \)
2. Applied the quotient rule: \( \frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)
3. Differentiated \( u \) and \( v \) with respect to x:
• For \( u \): \( \frac{du}{dx} = e^y \frac{dy}{dx} - e^x - y e^x \frac{dy}{dx} \)
• For \( v \): \( \frac{dv}{dx} = e^x - e^y - x e^y \frac{dy}{dx} \)
4. Substituted \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) back into the quotient rule formula and simplified.
This process helps to understand the complexity of how changes propagate through a function and its derivatives.