Problem 302 - Calculus Volume 1

Question

Use implicit differentiation to find \(\frac{d y}{d x}\). $$ x^{2} y=y-7 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{-2xy}{x^2 - 1} \)

Step 1: Differentiate both sides of the equation with respect to x

Start with the given equation: \( x^2 y = y - 7 \). Differentiate both sides with respect to \( x \), applying the product rule to \( x^2 y \). Remember, \( y \) is a function of \( x \). This gives us: \[ \frac{d}{dx}(x^2 y) = \frac{d}{dx}(y - 7) \].

Step 2: Apply the product rule to the left side

The product rule states that \( \frac{d}{dx}(u v) = u \frac{dv}{dx} + v \frac{du}{dx} \). In our case, \( u = x^2 \) and \( v = y \). Therefore, \[ \frac{d}{dx}(x^2 y) = x^2 \frac{dy}{dx} + y \cdot 2x \].

Step 3: Differentiate the right side of the equation

Differentiate the right-hand side of the equation: \( y - 7 \), treating \( y \) as a function of \( x \). The derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} \), and the derivative of \( -7 \) is 0, resulting in: \[ \frac{d}{dx}(y - 7) = \frac{dy}{dx} \].

Step 4: Substitute the derivatives back into the equation

Substitute the differentiated parts back into the equation: \[ x^2 \frac{dy}{dx} + 2xy = \frac{dy}{dx} \].

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

Isolate \( \frac{dy}{dx} \) by moving terms involving \( \frac{dy}{dx} \) to one side of the equation: \[ x^2 \frac{dy}{dx} - \frac{dy}{dx} = -2xy \]. Factor out \( \frac{dy}{dx} \): \[ (x^2 - 1)\frac{dy}{dx} = -2xy \]. Finally, divide both sides by \( x^2 - 1 \): \( \frac{dy}{dx} = \frac{-2xy}{x^2 - 1} \).

Key Concepts

Product Rule

One of the fundamental rules in calculus, especially when dealing with differentiation, is the product rule. The product rule is particularly useful when you are dealing with the derivative of a product of two functions. The rule states that if you have two functions, say \( u \) and \( v \), the derivative of their product \( u \, v \) with respect to \( x \) is given by: \[ \frac{d}{dx}(u v) = u \frac{dv}{dx} + v \frac{du}{dx} \] This means you take the derivative of the first function, multiply it by the second function, then add the first function times the derivative of the second function.

• Think of it as two parts:
  1. Derivative of the first function times the second function.
  2. Plus first function times derivative of the second function.

In our exercise, \( u = x^2 \) and \( v = y \). When we apply the product rule to \( x^2 y \), we get: \[ x^2 \frac{dy}{dx} + y \cdot 2x \] This is how we handle the left side of the equation through implicit differentiation.

Derivative

In calculus, the concept of a derivative is essential for understanding how functions change. The derivative provides a way to measure how a function's output changes as its input changes. For a function \( f(x) \), the derivative is often written as \( f'(x) \) or \( \frac{df}{dx} \). When a function is dependent on more than one variable or another function, like \( y(x) \), implicitly related to \( x \), the process is slightly different. This is where implicit differentiation comes into play.

• This technique differentiates both sides of an equation with respect to \( x \).
• It's crucial when \( y \) is not easily isolated.

In this exercise, we differentiate \( y - 7 \), knowing \( y \) is a function of \( x \). Therefore, the derivative \( \frac{d}{dx}(y - 7) = \frac{dy}{dx} \), since the derivative of a constant (like \(-7\)) is zero. This provides us with the necessary form of the equation to subsequently solve for \( \frac{dy}{dx} \).

Solve for dy/dx

The ultimate goal of implicit differentiation is to solve for \( \frac{dy}{dx} \), which represents the rate at which \( y \) changes concerning \( x \). In the final steps of these exercises, we aim to isolate \( \frac{dy}{dx} \) to see the relationship clearer. Initially, we substitute back the derivatives into the equation: \[ x^2 \frac{dy}{dx} + 2xy = \frac{dy}{dx} \] Next, we need to get all terms involving \( \frac{dy}{dx} \) on one side and the remaining terms on the other. This involves:

• Rearranging terms to isolate terms with \( \frac{dy}{dx} \)
• Factoring \( \frac{dy}{dx} \) out
• Dividing through by any remaining factors

Ultimately, this gives: \[ (x^2 - 1)\frac{dy}{dx} = -2xy \] And finally: \[ \frac{dy}{dx} = \frac{-2xy}{x^2 - 1} \] Completing these steps helps us understand and articulate how \( y \) changes as \( x \) does, following the implicit relation given.