Question

Implicit differentiation Use Theorem 15.9 to evaluate dy/dx. Assume each equation implicitly defines y as a differentiable function of \(x\). $$y e^{x y}-2=0$$

Short answer

Question: Find the derivative of the implicitly defined function \(y e^{x y}-2=0\) with respect to x. Answer: The derivative of the implicitly defined function with respect to x is given by \(\frac{dy}{dx} = \frac{- y' - y}{x}\).

Step-by-step solution

Differentiate both sides with respect to x

Differentiating both sides of the given equation with respect to x, we have: $$\frac{d}{dx}(ye^{xy}-2) = \frac{d}{dx}(0)$$

Apply Product Rule and Chain Rule

Recall the product rule: \((uv)' = u'v + uv'\). We apply the product rule to \(ye^{xy}\) and the chain rule for the exponential function. Differentiating with respect to x, we get: $$\frac{d}{dx}(y) \cdot e^{xy} + y \cdot \frac{d}{dx}(e^{xy}) - \frac{d}{dx}(2) = 0$$

Differentiate e^(xy) and Simplify

Now, we need to differentiate \(e^{xy}\) using the chain rule. Let \(u = xy\), then: $$\frac{d}{dx}(e^{xy}) = \frac{d}{dx}(e^u) \cdot \frac{d}{dx}(u)$$ Differentiate \(e^u\) with respect to \(u\) and \(u\) with respect to \(x\): $$\frac{d}{dx}(e^{xy}) = e^u \cdot (y + x \frac{dy}{dx}) = e^{xy}(y + x \frac{dy}{dx})$$ Substitute this result back into our previous equation: $$y'e^{xy} + y e^{xy}(y + x\frac{dy}{dx}) - 0 = 0$$

Solve for dy/dx

Pick out terms containing \(\frac{dy}{dx}\) on the left side and move the other terms to the other side: $$e^{xy}(y + x\frac{dy}{dx}) = - y'e^{xy} + 0$$ Now isolate \(\frac{dy}{dx}\) on one side: $$\frac{dy}{dx}(xe^{xy}) = - y'e^{xy} - e^{xy}y$$ Finally, divide both sides by \(xe^{xy}\): $$\frac{dy}{dx} = \frac{- y'e^{xy} - e^{xy}y}{xe^{xy}}$$

Simplify the Answer

Now, simplify the result by canceling the \(e^{xy}\) terms: $$\frac{dy}{dx} = \frac{- y' - y}{x}$$ So, the derivative of y with respect to x is given by: $$\frac{dy}{dx} = \frac{- y' - y}{x}$$

Key concepts

Product Rule

The Product Rule is a fundamental concept in calculus that is especially useful when differentiating functions that are the product of two other functions. In simple terms, if you have a function that is the product of two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is given by:

• \((uv)' = u'v + uv'\)

This means you take the derivative of the first function \( u \) while keeping the second function \( v \) constant, and then add it to the derivative of the second function \( v \) while keeping the first function \( u \) constant.
In the original exercise, applying the Product Rule to the term \( ye^{xy} \) involves taking the derivative of \( y \) times \( e^{xy} \) plus \( y \) times the derivative of \( e^{xy} \). This helps break down complex expressions into manageable parts for differentiation.

Chain Rule

The Chain Rule is a powerful tool in calculus used to differentiate composite functions. A composite function is one where a function is applied to the result of another function. When you have such a function in the form \( f(g(x)) \), the Chain Rule states the derivative is:

• \( f'(g(x)) \, g'(x) \)

This essentially means you first differentiate the outer function, keeping the inner function as it is, and then multiply it by the derivative of the inner function.
In the context of implicit differentiation, the Chain Rule is crucial for finding the derivative of \( e^{xy} \). You treat \( xy \) as the inner function and \( e^u \) (where \( u = xy \)) as the outer function. Differentiating accordingly helps in progressing towards finding \( \frac{dy}{dx} \).

Differentiable Function

A Differentiable Function is a function that can be differentiated everywhere within its domain. This implies that the function has a derivative at every point in its space. For implicit differentiation, it is assumed that the dependent variable is a differentiable function of the independent variable.
In the given exercise, we assume that \( y \) is implicitly defined as a differentiable function of \( x \). This means we can apply differentiation techniques like the Product Rule and Chain Rule to find \( \frac{dy}{dx} \). Keeping in view that the function needs to be smooth and continuous everywhere, ensures that our differentiation methods yield meaningful outputs.

Exponential Function

An Exponential Function is a mathematical function of the form \( e^x \), where \( e \) is Euler's number, approximately 2.718. Exponential functions have unique properties, especially their characteristic of their derivative being the same as the original function.

• The derivative of \( e^u \) is \( e^u \cdot u' \).

In the realm of calculus, they often appear in situations involving growth or decay due to their base, \( e \), being the natural base of logarithms.
In the provided exercise, the exponential function \( e^{xy} \) presents a scenario where you use the Chain Rule to handle differentiation because the exponent itself, \( xy \), is a function needing further differentiation. This highlights the intertwined nature of exponential and chain rules in complex calculus problems.