Question

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

Short 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-by-step solution

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.

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.

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 \(u'=\frac{16x^{2}+4x^{2}}{(3-x)^{2}} = \frac{20x^{2}}{(3-x)^{2}}\).

Substitute \(u'\) into \(y'\)

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

Simplify the Result

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