Question

1. if ( y=\frac{e^{x}}{3 x + 1} ), then ( \frac{d y}{d x}= )

(a) ( \frac{e^{x}(3 x - 2)}{(3 x + 1)^{2}} )
(b) ( \frac{e^{x}(3 x - 3)}{(3 x + 1)^{2}} )
(c) ( \frac{e^{x}(3 x + 2)}{(3 x + 1)^{2}} )
(d) ( \frac{e^{x}(3 x + 3)}{(3 x + 1)^{2}} )

Explanation:

Step1: Apply quotient rule

The quotient rule states that if \(y = \frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u = e^{x}\), \(u^\prime=e^{x}\); \(v = 3x + 1\), \(v^\prime=3\).

$$ \frac{dy}{dx}=\frac{e^{x}(3x + 1)-e^{x}\times3}{(3x + 1)^{2}} $$

Step2: Simplify the numerator

Factor out \(e^{x}\) from the numerator:

$$ \frac{e^{x}(3x + 1-3)}{(3x + 1)^{2}}=\frac{e^{x}(3x - 2)}{(3x + 1)^{2}} $$

Answer:

A. \(\frac{e^{x}(3x - 2)}{(3x + 1)^{2}}\)