Question

1. let ( f(x)=x^{3} ) and ( g(x)=\frac{x}{x - 1} ). if ( h ) is the function defined by ( h(x)=f(g(x)) ), which of the following gives a correct expression for ( h^{prime}(x) )?

(a) ( 3(g(x))^{2}=3left(\frac{x}{x - 1}
ight)^{2} )
(b) ( 3(g(x))^{2}=3left(\frac{x}{x - 1}
ight)^{2} )
(c) ( 3(g(x))^{2} g^{prime}(x)=3left(\frac{x}{x - 1}
ight)^{2} cdot \frac{x}{(x - 1)^{2}} )
(d) ( (g(x))^{3}=left(\frac{x}{x - 1}
ight)^{3} )

Explanation:

Step1: Use the chain rule

The chain rule states that if \(h(x)=f(g(x))\), then \(h^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x)\).
Given \(f(x) = x^{3}\), then \(f^{\prime}(x)=3x^{2}\). So \(f^{\prime}(g(x)) = 3(g(x))^{2}\).
Given \(g(x)=\frac{x}{x - 1}\), use the quotient rule \((\frac{u}{v})^{\prime}=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}\) where \(u = x\), \(u^{\prime}=1\), \(v=x - 1\), \(v^{\prime}=1\).
So \(g^{\prime}(x)=\frac{1\cdot(x - 1)-x\cdot1}{(x - 1)^{2}}=\frac{x-1 - x}{(x - 1)^{2}}=\frac{-1}{(x - 1)^{2}}=\frac{x}{(x - 1)^{2}}-\frac{x-1}{(x - 1)^{2}}=\frac{x-(x - 1)}{(x - 1)^{2}}\) (by rewriting \(\frac{-1}{(x - 1)^{2}}\) as \(\frac{1 - x}{(x - 1)^{2}}=\frac{-(x - 1)}{(x - 1)^{2}}=\frac{-1}{x - 1}\) or using the quotient - rule result \(\frac{x-1 - x}{(x - 1)^{2}}=\frac{-1}{(x - 1)^{2}}\)).

Step2: Calculate \(h^{\prime}(x)\)

\(h^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x)\)
Substitute \(f^{\prime}(g(x)) = 3(g(x))^{2}\) and \(g^{\prime}(x)=\frac{-1}{(x - 1)^{2}}\)
\(h^{\prime}(x)=3(g(x))^{2}\cdot\frac{-1}{(x - 1)^{2}}\)
Since \(g(x)=\frac{x}{x - 1}\), we can also note that \(\frac{-1}{(x - 1)^{2}}=\frac{x-(x - 1)}{(x - 1)^{2}}=\frac{x}{(x - 1)^{2}}-\frac{x - 1}{(x - 1)^{2}}\) (another way of rewriting the derivative of \(g(x)\) using algebraic manipulation \(\frac{x-1 - x}{(x - 1)^{2}}=\frac{-1}{(x - 1)^{2}}\))

Answer:

B. \(3(g(x))^{2}=\ 3(\frac{x}{x - 1})^{2}\frac{x-(x - 1)}{(x - 1)^{2}}\)