Alright, let's find the derivative of the function [tex]\( g(x) = \frac{e^x}{5-x} \)[/tex].
To do this, we'll use the quotient rule for differentiation. The quotient rule states that if you have a function in the form of [tex]\( \frac{u(x)}{v(x)} \)[/tex], its derivative is given by:
[tex]\[
\left(\frac{u(x)}{v(x)}\right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\][/tex]
Here, [tex]\( u(x) = e^x \)[/tex] and [tex]\( v(x) = 5 - x \)[/tex].
First, we need to find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
- The derivative of [tex]\( u(x) = e^x \)[/tex] is [tex]\( u'(x) = e^x \)[/tex].
- The derivative of [tex]\( v(x) = 5 - x \)[/tex] is [tex]\( v'(x) = -1 \)[/tex].
Now, applying the quotient rule:
[tex]\[
g'(x) = \frac{e^x \cdot (5 - x) - e^x \cdot (-1)}{(5 - x)^2}
\][/tex]
Simplify the numerator:
[tex]\[
g'(x) = \frac{e^x (5 - x) + e^x}{(5 - x)^2}
\][/tex]
Factor [tex]\( e^x \)[/tex] out of the numerator:
[tex]\[
g'(x) = \frac{e^x (5 - x + 1)}{(5 - x)^2}
\][/tex]
Simplify inside the parenthesis:
[tex]\[
g'(x) = \frac{e^x (6 - x)}{(5 - x)^2}
\][/tex]
Break it into two separate terms to make it clearer:
[tex]\[
g'(x) = e^x \left( \frac{6-x}{(5-x)^2} \right) = \frac{e^x}{5 - x} + \frac{e^x}{(5 - x)^2}
\][/tex]
So, the derivative of the function [tex]\( g(x) = \frac{e^x}{5-x} \)[/tex] is:
[tex]\[
g'(x) = \frac{e^x}{5 - x} + \frac{e^x}{(5 - x)^2}
\][/tex]
