Derivative of simple fraction

Question

What is the fastest way of finding the derivative of:

$\frac{x}{x+K}$ (simplified form)

is there a substitution I miss or does the quotient rule the job here? There should be a quick way of finding the derivative

Answer 1

Quotient rule works, although here's a probably slightly faster method.

Let $$\begin{align}y &= \dfrac{x}{x+K} \\\ln y &= \ln x - \ln(x+K) \\&\text{Differentiating w.r.t.} x \\\dfrac{1}{y}\dfrac{dy}{dx} &= \dfrac{1}{x} - \dfrac{1}{x+K} \\\dfrac{dy}{dx} &= y \left(\dfrac{1}{x} - \dfrac{1}{x+K}\right) \\ &= \dfrac{x}{x+K}\left(\dfrac{1}{x} - \dfrac{1}{x+K}\right) \\&= \dfrac{1}{x+K}-\dfrac{x}{(x+K)^2} \\&= \dfrac{x+K-x}{(x+K)^2} \\&= \dfrac{K}{(x+K)^2}\end{align}$$

Answer 2

by th quotient rule we get $y'=\frac{x+K-x}{(x+K)^2}=\frac{K}{(x+K)^2}$