Problem 66 - Calculus an Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=\left(\frac{4 x^{2}}{3-x}\right)^{3} $$

Step-by-Step Solution

Answer

The derivative of the function \( y = \left(\frac{4 x^{2}}{3-x}\right)^{3} \) is \( y' = 9 \left(\frac{4x^{2}}{3 - x} \right)^{3} \left(\frac{12x^{2} - 24x}{(3 - x)^{2}} \right) \).

Step 1: Apply Chain Rule

Applying the Chain Rule first, the derivative of \( y = (f(g(x)))^{3} \) becomes \( y' = 3(f(g(x)))^{2} f'(g(x)) g'(x) \), where \( f(x) = x^{3} \), \( g(x) = \frac{4x^{2}}{3 - x} \), \( f'(x) = 3x^{2} \), and \( g'(x) = \frac{d}{dx}(\frac{4x^{2}}{3 - x}) \). Now, the task is to find \( g'(x) \).

Step 2: Apply Quotient Rule

Applying the Quotient Rule to find \( g'(x) \), the derivative \( h'(x) \) of a function of the form \( h(x) = \frac{p(x)}{q(x)} \) is \( h'(x) = \frac{p'(x)q(x) - p(x)q'(x)}{[q(x)]^{2}} \). Here, \( p(x) = 4x^{2} \), \( q(x) = 3 - x \), \( p'(x) = 8x \), and \( q'(x) = -1 \). Substituting these into the Quotient Rule gives \( g'(x) = \frac{8x(3 - x) - 4x^{2}(-1)}{(3 - x)^{2}} \). Simplifying this gives \( g'(x) = \frac{8x^{2} - 24x + 4x^{2}}{(3 - x)^{2}} = \frac{12x^{2} - 24x}{(3 - x)^{2}} \).

Step 3: Substitute back

Substitute \( f'(g(x)) g'(x) \) back into Step 1 to find the derivative of the original function: \( y' = 3(f(g(x)))^{2} f'(g(x)) g'(x) = 3 \left(\frac{4x^{2}}{3 - x} \right)^{2} * 3 \left(\frac{4x^{2}}{3 - x} \right) * \left(\frac{12x^{2} - 24x}{(3 - x)^{2}} \right) = 9 \left(\frac{4x^{2}}{3 - x} \right)^{3} \left(\frac{12x^{2} - 24x}{(3 - x)^{2}} \right) \)

Key Concepts

Derivative of a Function

Understanding the concept of a derivative is fundamental to calculus. Every time you hear the term 'derivative', think about it as a way to express the rate at which one quantity changes with respect to another. In the context of functions, finding the derivative—often denoted by the symbols \' \( f'(x) \) \' or \' \( y' \) \'—means determining how the function's output changes as its input changes.

When we calculate the derivative of a function, we are essentially looking for the slope of the function at any point along its curve. This slope indicates the steepness and direction of a line tangent to the function's graph at that point. In the world of motion, for example, if the function represents distance over time, the derivative would be the velocity, telling us how fast the position is changing at any given moment.

Derivatives are the backbone of differential calculus and have a wealth of applications in science, engineering, economics, and beyond because they allow us to predict and understand behavior in a precisely quantitative way.

Chain Rule

The chain rule is a powerful tool in calculus used for finding the derivative of a composite function. Imagine you're wearing a chain made of many interlinked parts; the chain rule works similarly by breaking down the process into simpler parts that are easier to manage.

When faced with a composite function, which is essentially a function within another function—like a nesting doll— the chain rule instructs us to differentiate the outer function first, keeping the inner function untouched. Then, we multiply this by the derivative of the inner function.

For instance, if we have a composite function \' \( y=(f(g(x)))^3 \) \' where \' \( f(x)=x^3 \) \' and \' \( g(x) \) \' is another function of \' \( x \) \', then the derivative of \' \( y \) \' is \' \( y'=3(f(g(x)))^2 f'(g(x))g'(x) \) \' as shown in our exercise. The chain rule is used here because the function being raised to a power is itself another function, thus creating the necessity for a 'chain' of derivatives.

Quotient Rule

The quotient rule is a derivative rule that allows us to find the slope (derivative) of ratios of functions. It's especially helpful when we have one function divided by another—hence 'quotient,' which is another word for division.

To apply the quotient rule, we start with a function that can be expressed as \' \( h(x) = \frac{p(x)}{q(x)} \) \', where \' \( p(x) \) \' and \' \( q(x) \) \' are both functions of \' \( x \) \' themselves. According to the quotient rule, the derivative of \' \( h(x) \) \' is \' \( h'(x) = \frac{p'(x)q(x) - p(x)q'(x)}{\left[q(x)\right]^2} \) \' which is essentially the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all over the square of the denominator.

When diving into the details of our exercise, where \' \( g(x) = \frac{4x^2}{3 - x} \) \', we utilize the quotient rule to find \' \( g'(x) \) \' by differentiating the numerator \' \( 4x^2 \) \' and denominator \' \( 3 - x \) \' separately, then applying the rule to achieve \' \( g'(x) = \frac{8x(3 - x) - 4x^2(-1)}{(3 - x)^2} \) \' and simplifying to obtain the final derivative. Through this method, we manage to express how the rate of change for the ratio of these two functions behaves.