Question

differentiate.
$f(x) = \frac{5}{(2x - 3)^4}$
$\bigcirc$ $f(x) = \frac{5}{4(2x - 3)^3}$
$\bigcirc$ $f(x) = \frac{-40}{(2x - 3)^3}$
$\bigcirc$ $f(x) = \frac{5}{8(2x - 3)^5}$
$\bigcirc$ $f(x) = \frac{-40}{(2x - 3)^5}$

Explanation:

Step1: Rewrite the function

Rewrite \( f(x)=\frac{5}{(2x - 3)^4} \) as \( f(x)=5(2x - 3)^{-4} \).

Step2: Apply the chain rule

The chain rule states that if \( y = u^n \) and \( u = g(x) \), then \( y'=n u^{n - 1}\cdot u' \). Here, \( u = 2x - 3 \), \( n=-4 \), and the derivative of \( u \) with respect to \( x \) is \( u' = 2 \).
First, find the derivative of the outer function: \( \frac{d}{du}(5u^{-4})=5\times(-4)u^{-5}=-20u^{-5} \).
Then, multiply by the derivative of the inner function: \( -20(2x - 3)^{-5}\times2=-40(2x - 3)^{-5} \).
Rewrite \( -40(2x - 3)^{-5} \) as \( \frac{-40}{(2x - 3)^5} \).

Answer:

\( f'(x)=\frac{-40}{(2x - 3)^5} \) (the fourth option)