91Ó°ÊÓ

The chain rule is a fundamental principle in calculus that allows us to find the derivative of a composite function. When dealing with functions embedded within other functions, the chain rule comes in handy. It means you take the derivative of the outer function and multiply it by the derivative of the inner function.

Here’s how it works: Suppose you have a function where one function is inside another, like \( y = f(g(x)) \). The derivative of \( y \) with respect to \( x \) is expressed as \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). This expression helps us to find how much \( y \) changes when \( x \) changes.

In the given exercise, we applied the chain rule to find the derivative of \( y = f\left(\frac{2x+3}{3-2x}\right) \). By first finding the derivative of the inside function \( \frac{2x+3}{3-2x} \) as \( \frac{du}{dx} \), then multiplying it by the derivative of the outside function \( f'(u) \), we effectively used the chain rule to solve the problem.

• The inside function \( u = \frac{2x+3}{3-2x} \) had its derivative calculated using the quotient rule.
• The outer function \( f'(u) = \sin(\log u) \) determined the overall change in \( y \) for a small change in \( u \).
• Overall, \( \frac{dy}{dx} = f'(u) \cdot \frac{du}{dx} \) was the application of the chain rule.

The quotient rule is an essential technique in calculus for finding the derivative of a quotient of two functions. This rule is extremely useful when both the numerator and the denominator are functions of the variable, like the function given in our example: \( u = \frac{2x+3}{3-2x} \).

The quotient rule states: If \( u = \frac{g(x)}{h(x)} \), then the derivative \( \frac{du}{dx} \) is given by:\[ \frac{du}{dx} = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]To apply the quotient rule effectively, just remember:

• Differentiate the top function, the numerator \( g(x) \), to get \( g'(x) \).
• Differentiate the bottom function, the denominator \( h(x) \), to get \( h'(x) \).
• Plug these into the formula, ensuring you subtract \( g(x) \) times \( h'(x) \) from \( g'(x)h(x) \).
• Divide everything by the square of the denominator \( (h(x))^2 \).

Our exercise used this rule perfectly. We identified \( g(x) = 2x + 3 \) and \( h(x) = 3 - 2x \):

• \( g'(x) = 2 \)
• \( h'(x) = -2 \)

Then, we plugged these into the formula to find \( \frac{du}{dx} = \frac{12}{(3 - 2x)^2} \). This calculated derivative allowed us to apply the chain rule for solving the original problem.

Trigonometric functions such as \( \sin \), \( \cos \), and \( \log \) often feature prominently in calculus, especially in composite functions. Whenever they are involved, understanding their properties and derivatives is crucial to solving calculus problems efficiently.

In this exercise, the function \( y = f\left(\frac{2x+3}{3-2x}\right) \) involved the derivative of a trigonometric and logarithmic composite function, \( f'(x) = \sin(\log x) \). Here's a brief overview:

• The sine function \( \sin(x) \) and its derivative is \( \cos(x) \). Derivatives of trigonometric functions follow specific rules and can require additional techniques like the chain rule.
• The logarithm function \( \log(x) \) is a component of our trigonometric function here. Knowing that the derivative of \( \log(x) \) is \( \frac{1}{x} \) helps in understanding how \( \sin(\log x) \) behaves.
• Combining these functions complicates derivatives but also showcases the power and necessity of rules like the chain and quotient rules to solve such problems.

In our step-by-step solution, we saw that after calculating \( \frac{du}{dx} \) with the quotient rule, the solution required substituting \( u = \frac{2x+3}{3-2x} \) back to get \( \sin(\log(\frac{2x+3}{3-2x})) \) in the derivative formula. This effectively utilized trigonometric and logarithmic properties to get the correct derivative \( \frac{dy}{dx} = \frac{12}{(3-2x)^2} \sin(\log(\frac{2x+3}{3-2x})) \).