Question

Differentiate. bek y = 2ex + 1 ex (2ex + 1)2 O 6ex (2x + 1)2 6ex (2x + 1)3 6ex (2e +1)

Answer 1

Given function

$y=\frac{6e^x}{2e^x+1}$

Differentiate y with respect to x,

$\frac{dy}{dx}=\frac{d(\frac{6e^x}{2e^x+1})}{dx}$

First take the constant out from numerator,

$\frac{dy}{dx}=6\frac{d(\frac{e^x}{2e^x+1})}{dx}$

Apply the quotient rule of differentiateion,

$(\frac{f}{g})^{\prime}=\frac{fg^{\prime}-gf^{\prime}}{g^2}$

$\begin{gathered} \frac{dy}{dx}=6\frac{\frac{de^x}{dx}(2e^x+1)-e^x\frac{d(2e^x+1)}{dx}^{}^{}}{(2e^x+1)^2} \\ \frac{dy}{dx}=6\frac{e^x(2e^x+1)-e^x2e^x^{}}{(2e^x+1)^2} \\ \frac{dy}{dx}=6\frac{2e^{2x}+e^x-2e^{2x}}{(2e^x+1)^2} \\ \frac{dy}{dx}=\frac{6e^x}{(2e^x+1)^2} \end{gathered}$

The derivative of the given function is (b). 6e^x/(2e^x+1)^2

ANSWER : B