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=(6e^x)/(2e^x+1)`

Differentiate y with respect to x,

`(dy)/(dx)=(d((6e^x)/(2e^x+1)))/(dx)`

First take the constant out from numerator,

`(dy)/(dx)=6(d((e^x)/(2e^x+1)))/(dx)`

Apply the quotient rule of differentiateion,

`((f)/(g))^(\prime)=(fg^(\prime)-gf^(\prime))/(g^2)`

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

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

ANSWER : B