16. if ( y=\frac{e^{x}}{3 x + 1} ), then ( \frac{d y}{d x}= )\n(a) ( \frac{e^{x}(3 x - 2)}{(3 x + 1)^{2}}…

16. if ( y=\frac{e^{x}}{3 x + 1} ), then ( \frac{d y}{d x}= )\n(a) ( \frac{e^{x}(3 x - 2)}{(3 x + 1)^{2}} )\n(b) ( \frac{e^{x}(3 x - 3)}{(3 x + 1)^{2}} )\n(c) ( \frac{e^{x}(3 x + 2)}{(3 x + 1)^{2}} )\n(d) ( \frac{e^{x}(3 x + 3)}{(3 x + 1)^{2}} )

16. if ( y=\frac{e^{x}}{3 x + 1} ), then ( \frac{d y}{d x}= )\n(a) ( \frac{e^{x}(3 x - 2)}{(3 x + 1)^{2}} )\n(b) ( \frac{e^{x}(3 x - 3)}{(3 x + 1)^{2}} )\n(c) ( \frac{e^{x}(3 x + 2)}{(3 x + 1)^{2}} )\n(d) ( \frac{e^{x}(3 x + 3)}{(3 x + 1)^{2}} )

Answer

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}})