Problem 8 - Calculus for Biology and Medicine

Question

In Problems \(1-8\), find \(\frac{d y}{d x}\) by implicit differentiation. $$ \frac{x}{x y+1}=2 x y $$

Step-by-Step Solution

Answer

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

Step 1: Differentiate both sides with respect to x

Begin by differentiating both sides of the equation \( \frac{x}{xy + 1} = 2xy \) with respect to \( x \). We use the quotient rule for the left side and the product rule for the right side.

Step 2: Use the Quotient Rule on the Left Side

The left side of the equation is \( \frac{x}{xy + 1} \). By the quotient rule, \( \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} \). Let \( f(x) = x \) and \( g(x) = xy + 1 \). Then, \( \frac{d}{dx}\left(\frac{x}{xy + 1}\right) = \frac{1(xy + 1) - x(\frac{d}{dx}(xy + 1))}{(xy + 1)^2} \).

Step 3: Differentiate xy+1 with respect to x on the Left Side

Now, differentiate \( xy + 1 \) to find \( \frac{d}{dx}(xy + 1) \). Using the product rule for \( xy \), this yields \( y + x\frac{dy}{dx} \). Substituting back, the differentiation becomes \[ \frac{(xy + 1) - x(y + x\frac{dy}{dx})}{(xy + 1)^2}. \]

Step 4: Use the Product Rule on the Right Side

Differentiate the right side \( 2xy \) with respect to \( x \) using the product rule. The derivative is \( 2(y + x\frac{dy}{dx}) \).

Step 5: Set Derivatives Equal and Simplify

Now equate the derivatives of both sides obtained from Steps 2 and 4:\[ \frac{(xy + 1) - x(y + x\frac{dy}{dx})}{(xy + 1)^2} = 2(y + x\frac{dy}{dx}). \]Simplify and solve for \( \frac{dy}{dx} \).

Step 6: Solve for dy/dx

Simplify both sides to isolate \( \frac{dy}{dx} \). After clearing fractions and simplifying, rearrange to get:\[ \frac{xy + 1 - xy - x^2\frac{dy}{dx}}{(xy + 1)^2} = 2y + 2x\frac{dy}{dx}, \]\[ \frac{1 - x^2\frac{dy}{dx}}{(xy + 1)^2} = 2y + 2x\frac{dy}{dx}. \]

Step 7: Simplify to Find dy/dx

Multiply through by \( (xy + 1)^2 \) to clear the fraction:\[ 1 - x^2\frac{dy}{dx} = (2y + 2x\frac{dy}{dx})(xy + 1)^2. \]Rearrange terms and solve for \( \frac{dy}{dx} \), yielding:\[ \frac{dy}{dx} = \frac{1 - 2y(xy + 1)^2}{2x(xy + 1)^2 - x^2}, \] assuming algebraic simplification and evaluation.

Key Concepts

Quotient Rule

When you're working with differentiation and need to differentiate a function expressed as a quotient of two other functions, you'll be using the quotient rule. This is essential when tackling problems where one function is divided by another. The rule itself can be a bit tricky at first, but with practice, it becomes a powerful tool.

To apply the quotient rule, consider the function as the ratio of two functions: \( \frac{f(x)}{g(x)} \). The quotient rule states that the derivative of \( \frac{f}{g} \) is:

• \( \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} \)

Here, \( f'(x) \) and \( g'(x) \) are the derivatives of \( f(x) \) and \( g(x) \), respectively.

Using this rule, you find the derivative by first calculating the derivative of the numerator \( f(x) \) and the denominator \( g(x) \), plugging them into the formula accordingly.

Keep in mind:- Always square the denominator for the result.- Careful calculation of each derivative is crucial.- Simplify your final expression to make it easier to interpret.

Product Rule

The product rule is a fundamental technique used when you differentiate a product of two functions. This might pop up often where two functions are multiplied together, like in our given problem.

When two functions are multiplied, say \( u(x) \) and \( v(x) \), the product rule allows us to find their derivative. It is expressed as:

• \( \frac{d}{dx}(uv) = u'v + uv' \)

Here, \( u'(x) \) and \( v'(x) \) are the derivatives of \( u(x) \) and \( v(x) \), respectively.

The approach is straightforward: differentiate each function individually, then multiply and add as specified by the rule.

Remember:- Perfectly align the terms in the formula.- Practice makes it part of your muscle memory, making it quick and efficient.

Derivative of a Function

At the heart of calculus, taking the derivative of a function is a core operation. It lets us understand how a function changes—both in its direction and steepness.

When performing implicit differentiation, as with our exercise, we must carefully differentiate each term regarding the desired variable while considering dependencies shown through related terms.

Implicit differentiation usually comes in handy when functions aren’t explicitly solved for one variable in terms of another. This means treating one variable as implicitly dependent on another, such as \( y \) implicitly in terms of \( x \). Steps typically involve:

• Differentiating both sides of an equation.
• Applying derivative rules like product or quotient rules whenever necessary.
• Solving for the desired derivative, \( \frac{dy}{dx} \), in our case.

These steps will allow us to discover the rate of change concerning the constraints expressed in the function, aiding in interpretation and further calculations.