Question

In Exercises, find \(d y / d x\) implicitly. $$ e^{x y}+x^{2}-y^{2}=10 $$

Short answer

The derivative of the function, \(\frac{dy}{dx}\), is \(\frac{dy}{dx} = \frac{2x - e^{xy}}{e^{xy}x + 2y}\)

Step-by-step solution

Differentiate Both Sides

Differentiate both sides of the given equation with respect to \(x\). The derivative of \(e^{xy}\) is \(e^{xy}(y+x \frac{dy}{dx})\), using the chain rule. This rule permits us to differentiate a composite function. The derivative of \(x^2\) is \(2x\), and for \(-y^2\) it's \(-2y\frac{dy}{dx}\) by chain rule.

Set Derivation Equal to Zero

We set the derived equation equal to zero, as this will enable us to solve for \(\frac{dy}{dx}\)

Solve for dy/dx

Now by isolating the terms involving \(\frac{dy}{dx}\), we can express \(\frac{dy}{dx}\) as a function of \(x\) and \(y\).

Key concepts

Chain Rule

The Chain Rule is an essential tool in calculus, particularly when working with functions of functions, also known as composite functions. When you differentiate composite functions, the Chain Rule helps manage the complexity of these combined expressions. For instance, if you have a function that consists of an exponent such as \(e^{xy}\), it suggests that there is a function within another function.
The Chain Rule states that if you have a composite function \(f(g(x))\), you must take the derivative of the outer function, \(f\), evaluated at the inner function, \(g(x)\), and multiply it by the derivative of the inner function, \(g'(x)\). In symbolic form, it looks like this:

• \(f'(g(x)) \cdot g'(x)\)

In the given problem, we applied the Chain Rule to the term \(e^{xy}\). Here, \(e\) is the outer function and \(xy\) is the inner function. We first differentiated the outer function with respect to \(xy\), keeping \(e^{xy}\) intact, and multiplied it by the derivative of the inner function \(xy\), considering \(x\) as a constant when differentiating with respect to \(y\). This results in a more manageable equation, aiding in finding the derivative \(\frac{dy}{dx}\).

Differentiation

Differentiation is the process used in calculus to find the rate at which a function is changing at any given point, which essentially gives you the slope of the function at that instant. It's a critical operation used to calculate derivatives, which are fundamental in analyzing functions.
When executing differentiation, each term is considered separately. In the problem, we use implicit differentiation, a method where we differentiate both sides of the equation concerning a variable \(x\), even if \(y\) is also involved as a function of \(x\).

• The derivative of \(x^2\) becomes \(2x\), since it's a straightforward power function.
• When differentiating \(-y^2\), you should take \(2y\frac{dy}{dx}\) because \(y\) itself is a function of \(x\).

Implicit differentiation helps to find \(\frac{dy}{dx}\) without having to explicitly solve \(y\) in terms of \(x\), which is useful when \(y\) is bound within a composite or complex structure.
This allows you to deal neatly with equations where \(y\) is not isolated.

Composite Function

A composite function is essentially a combination of two or more functions. The output of one function becomes the input of another function, creating layers of operations that must be navigated through techniques like the Chain Rule. When you have a function inside another function, such as \(e^{xy}\), it indicates a composite structure.
Breaking down composite functions requires understanding each layer thoroughly, whereby each part must be differentiated respectably.

• Identify the outer and inner functions. For \(e^{xy}\), \(e\) is the outer layer, while \(xy\) is processed inside it.
• Apply the Chain Rule to differentiate, first dealing with the outer function \(e\) while maintaining the inner function. Then, the derivative of the inner function \(xy\) is computed.

Composite functions can integrate various mathematical expressions, increasing complexity in differentiation. But with a clear approach, breaking these down into simpler function layers means you can adeptly handle derivatives and any implicit relationships between variables, like \(x\) and \(y\). This understanding simplifies complex equations into solvable expressions for \(\frac{dy}{dx}\).