Answer:
[tex]y' = \frac{1 - 2x}{e^{4x}}[/tex]
Step-by-step explanation:
If we have a quotient function y, in the following format
[tex]y = \frac{f(x)}{g(x)}[/tex]
This function has the following derivative
[tex]y' = \frac{f'(x)*g(x) - g'(x)*f(x)}{(g(x))^{2}}[/tex]
In this problem, we have that:
[tex]f(x) = x, g(x) = e^{2x}[/tex]
So
[tex]f'(x) = 1, g'(x) = 2e^{2x}[/tex]
The derivative of the function is:
[tex]y' = \frac{f'(x)*g(x) - g'(x)*f(x)}{(g(x))^{2}}[/tex]
[tex]y' = \frac{e^{2x} - 2xe^{2x}}{(e^{2x})^{2}}[/tex]
[tex]y' = \frac{e^{2x}(1 - 2x)}{e^{4x}}[/tex]
[tex]y' = \frac{1 - 2x}{e^{4x}}[/tex]
