Problem 66 - College Algebra and 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 \(y=(\frac{4 x^{2}}{3-x})^{3}\) is \(y' = \frac{240x^{4}}{(3-x)^{4}}\). The rules used to find this derivative were the power rule, the chain rule, and the quotient rule.

Step 1: Identify the Inner Function

The given function is \(y=(\frac{4 x^{2}}{3-x})^{3}\). Notice that the function inside the brackets, \(u=\frac{4 x^{2}}{3-x}\), can be considered as an inner function which has been raised to the power of 3.

Step 2: Apply the Chain Rule

To start, the outer function will be differentiated with respect to this inner function u. According to the chain rule, this gives \(y'=3u^{2}\cdot u'\). Here \(u'\) is the derivative of the inner function, which we will compute in the next step.

Step 3: Differentiate the Inner Function Using Quotient Rule

To find \(u'\), apply the quotient rule to the inner function \(u=\frac{4 x^{2}}{3-x}\). The quotient rule states that the derivative of \(\frac{p}{q}\) is \(\frac{q\cdot p' - p\cdot q'}{q^{2}}\). Applying the rule to the function gives \(u'=\frac{(3-x)\cdot 8x - (4x^{2})\cdot (-1)}{(3-x)^{2}}\). Simplifying this expression gives \(u'=\frac{16x^{2}+4x^{2}}{(3-x)^{2}} = \frac{20x^{2}}{(3-x)^{2}}\).

Step 4: Substitute \(u'\) into \(y'\)

Substitute \(u'\) and u into the expression for \(y'\) found in Step 2. This gives \(y' = 3(\frac{4 x^{2}}{3-x})^{2}\cdot \frac{20x^{2}}{(3-x)^{2}}\)

Step 5: Simplify the Result

Simplify the expression to get the derivative of the function. This yields \(y' = \frac{240x^{4}}{(3-x)^{4}}\).