Question

Find the derivatives of the following by first taking the natural log of both sides: $$(a) y=\frac{3 x}{(x+2)(x+4)}$$ $$(b) y=\left(x^{2}+3\right) e^{x^{2}+1}$$

Short answer

(a) \( \frac{dy}{dx} = \frac{3((x+2)(x+4) - x(x+4) - x(x+2))}{x(x+2)^2(x+4)^2}\)\; (b) \( \frac{dy}{dx} = 2x e^{x^2+1} (1 + x^2+3) \)

Step-by-step solution

Take the natural log of both sides (Part a)

First, we start with \( y = \frac{3x}{(x+2)(x+4)} \). Take the natural logarithm of both sides. This gives us \( \ln(y) = \ln\left(\frac{3x}{(x+2)(x+4)}\right) \). Using logarithm properties, we can expand this to \( \ln(y) = \ln(3) + \ln(x) - \ln(x+2) - \ln(x+4) \).

Differentiate both sides with respect to x (Part a)

Differentiate \( \ln(y) \) with respect to \( x \) using implicit differentiation. \( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x} - \frac{1}{x+2} - \frac{1}{x+4} \).

Solve for \( \frac{dy}{dx} \) (Part a)

Multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \). We have \( \frac{dy}{dx} = y \left( \frac{1}{x} - \frac{1}{x+2} - \frac{1}{x+4} \right) \). Substitute \( y = \frac{3x}{(x+2)(x+4)} \) back into the equation to get \( \frac{dy}{dx} = \frac{3x}{(x+2)(x+4)} \left( \frac{1}{x} - \frac{1}{x+2} - \frac{1}{x+4} \right) \).

Take the natural log of both sides (Part b)

Now we handle \( y = (x^2+3) e^{x^2+1} \). Take the natural log of both sides, obtaining \( \ln(y) = \ln((x^2+3) e^{x^2+1}) \). Simplify using logarithm properties: \( \ln(y) = \ln(x^2+3) + (x^2+1) \).

Differentiate both sides with respect to x (Part b)

Differentiate with respect to \( x \). For the left side: \( \frac{1}{y} \cdot \frac{dy}{dx} \). For the right side: derivative of \( \ln(x^2+3) \) is \( \frac{2x}{x^2+3} \), and derivative of \( x^2+1 \) is \( 2x \). This gives: \( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{2x}{x^2+3} + 2x \).

Solve for \( \frac{dy}{dx} \) (Part b)

To find \( \frac{dy}{dx} \), multiply both sides by \( y \). \( \frac{dy}{dx} = y \left( \frac{2x}{x^2+3} + 2x \right) \). Substitute \( y = (x^2+3) e^{x^2+1} \) back in: \( \frac{dy}{dx} = (x^2+3) e^{x^2+1} \left( \frac{2x}{x^2+3} + 2x \right) \).

Simplify the derivative expression (Part b)

Distribute \( (x^2+3) e^{x^2+1} \) with \( \left( \frac{2x}{x^2+3} + 2x \right) \). This results in \( \frac{dy}{dx} = 2x e^{x^2+1} + 2x(x^2+3) e^{x^2+1} \). Factor if needed or leave in expanded form.

Key concepts

Natural Logarithms

Natural logarithms (often shortened to "ln") are a type of logarithm where the base is the mathematical constant \( e \), approximately equal to 2.71828. Understanding how natural logs work is crucial in calculus, especially when tackling problems that involve exponentials or products/quotients of functions.

One of the main properties of logarithms, including natural logs, is that they convert multiplication into addition, and division into subtraction. This means if you have a logarithm of a product, you can break it down into a sum of logarithms. Similarly, a division inside a log can be expressed as a subtraction of logs. For example:

\[ \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \]

This property allows us to simplify complex expressions and is used extensively in calculus to make differentiation easier. In particular, taking the natural logarithm of both sides of an equation can help facilitate differentiation, as it did in the exercise step-by-step solution above.

Taking logs can also make it easier to apply derivative rules, as it transforms problems involving products and quotients into linear sums and differences, simplifying the process of finding derivatives.

Implicit Differentiation

Implicit differentiation is a powerful method used when dealing with equations that aren't easily solvable for one variable in terms of another. It's a different approach compared to explicit differentiation, where one variable is isolated on one side of the equation.

In a situation where an equation defines a relationship like \( y = \frac{3x}{(x+2)(x+4)} \), implicit differentiation allows us to differentiate both sides of this relationship with respect to \( x \), even if \( y \) is not isolated.

Here’s how it works: you assume both sides of the equation are functions of \( x \). You then take the derivative with respect to \( x \), treating \( y \) as an implicit function of \( x \). In the exercise, for instance, \( \ln(y) = \ln(3) + \ln(x) - \ln(x+2) - \ln(x+4) \), implicit differentiation applies the chain rule, giving \[ \frac{1}{y} \cdot \frac{dy}{dx} \] for the left side, and differentiating each term on the right accordingly.

The key to implicit differentiation is remembering to take the derivative of every instance of \( y \) with respect to \( x \) by always multiplying by \( \frac{dy}{dx} \). This ensures the change of \( y \) with respect to the changes in \( x \) is taken into account.

Derivative Rules

Derivative rules form the foundation for finding derivatives and help break down complex differentiation tasks into simpler, more manageable parts. Some fundamental rules used in the exercise are:

• Product Rule: Used when differentiating a product of two functions. If \( f(x) = u \, v \), where \( u \) and \( v \) are functions of \( x \), then \( f'(x) = u' \, v + u \, v' \).
• Quotient Rule: Necessary when differentiating a division of two functions. If \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{u' \, v - u \, v'}{v^2} \).
• Chain Rule: This rule is used when differentiating a function of a function, like \( e^{x^2+1} \). It states that if you have a composite function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).

In both parts of the exercise solution above, these rules are applied. For example, when deriving \( y = (x^2+3) e^{x^2+1} \), the use of the product rule first, followed by the chain rule for differentiating \( e^{x^2+1} \), demonstrates how these rules come together to solve a complex differentiation problem effectively.

Mastering these derivative rules is essential for performing advanced calculus tasks and solving a wide range of mathematical problems encountered in calculus.