To find the derivative of the function [tex]\( g(x) = \frac{e^x}{2 - 4x} \)[/tex], we can apply the quotient rule of differentiation. The quotient rule states that if we have a function that is the quotient of two functions, [tex]\( g(x) = \frac{f(x)}{h(x)} \)[/tex], then its derivative is given by:
[tex]\[
g'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{(h(x))^2}
\][/tex]
For our function [tex]\( g(x) = \frac{e^x}{2 - 4x} \)[/tex], we can identify [tex]\( f(x) = e^x \)[/tex] and [tex]\( h(x) = 2 - 4x \)[/tex].
First, we need to find the derivatives of [tex]\( f(x) \)[/tex] and [tex]\( h(x) \)[/tex]:
[tex]\[
f(x) = e^x \quad \Rightarrow \quad f'(x) = e^x
\][/tex]
[tex]\[
h(x) = 2 - 4x \quad \Rightarrow \quad h'(x) = -4
\][/tex]
Using the quotient rule:
[tex]\[
g'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{(h(x))^2}
\][/tex]
Substitute [tex]\( f(x) \)[/tex], [tex]\( f'(x) \)[/tex], [tex]\( h(x) \)[/tex], and [tex]\( h'(x) \)[/tex] into the quotient rule formula:
[tex]\[
g'(x) = \frac{e^x (2 - 4x) - e^x (-4)}{(2 - 4x)^2}
\][/tex]
Simplify the numerator:
[tex]\[
g'(x) = \frac{e^x (2 - 4x) + 4e^x}{(2 - 4x)^2}
\][/tex]
Factor out [tex]\( e^x \)[/tex] from the numerator:
[tex]\[
g'(x) = \frac{e^x (2 - 4x + 4)}{(2 - 4x)^2}
\][/tex]
Simplify inside the parentheses:
[tex]\[
g'(x) = \frac{e^x (6 - 4x)}{(2 - 4x)^2}
\][/tex]
Rewrite the expression:
[tex]\[
g'(x) = \frac{(6 - 4x)e^x}{(2 - 4x)^2}
\][/tex]
We can separate the two terms inside the numerator:
[tex]\[
g'(x) = \frac{6e^x}{(2 - 4x)^2} - \frac{4xe^x}{(2 - 4x)^2}
\][/tex]
These terms combine to give:
[tex]\[
g'(x) = \frac{e^x}{2 - 4x} + \frac{4e^x}{(2 - 4x)^2}
\][/tex]
Thus, the derivative of the function [tex]\( g(x) = \frac{e^x}{2 - 4x} \)[/tex] is:
[tex]\[
g'(x) = \frac{e^x}{2 - 4x} + \frac{4e^x}{(2 - 4x)^2}
\][/tex]
