Answer:
[tex]\displaystyle\frac{d}{dp}\bigg[ \displaystyle\frac{1}{p + ke^p}\bigg] =-\displaystyle\frac{ ke^p + 1}{(p + ke^p)^2}[/tex]
Step-by-step explanation:
We are given the function:
[tex]\displaystyle\frac{1}{p + ke^p}[/tex]
where k is a constant.
We use quotient rule to differentiate the given function
[tex]\bigg(\displaystyle\frac{f}{g}\bigg)' = \displaystyle\frac{gf' =- fg'}{g^2}[/tex]
Using the quotient rule and differentiating, we get:
[tex]\displaystyle\frac{d}{dp}\bigg[ \displaystyle\frac{1}{p + ke^p}\bigg]\\\\= \displaystyle\frac{(p+ke^p)(1)' - (1)(p + ke^p)'}{(p + ke^p)^2}\\\\= \displaystyle\frac{(p+ke^p)(0) - (1)(1 + ke^p)}{(p + ke^p)^2}\\\\= \displaystyle\frac{-(1 + ke^p)}{(p + ke^p)^2}\\\\=-\displaystyle\frac{ ke^p + 1}{(p + ke^p)^2}[/tex]
