Question

Find the derivative of the function at the given number.

$f\left(x\right)=\frac{x}{2-x},at-3$.

Short answer

The derivative of the function at -3is$\frac{2}{25}$.

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=\frac{x}{2-x},at-3$

Step 2. Formula used.

The derivative of a function f at a number,denoted by$f\text{'}\left(a\right),$is

.$f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative, with,$a=-3$ we have

$\begin{array}{c}f\text{'}(-3)=\underset{h\to 0}{\lim }\frac{f(-3+h)-f(-3)}{h}Definitionoff\text{'}(-3)\\ =\underset{h\to 0}{\lim }\frac{\left[\frac{(-3+h)}{2-(-3+h)}\right]-\left[\frac{-3}{2+3}\right]}{h}f\left(x\right)=\frac{x}{2-x}\\ =\underset{h\to 0}{\lim }\frac{\frac{(-3+h)}{5-h}+\frac{3}{5}}{h}Expand\\ =\underset{h\to 0}{\lim }\frac{\frac{-15+5h+15-3h}{5(h+5)}}{h}\\ =\underset{h\to 0}{\lim }\frac{2h}{5h(h+5)}Simplify\\ =\underset{h\to 0}{\lim }\frac{2}{5(h+5)}Cancelh\\ =\frac{2}{25}\end{array}$

Therefore, the derivative of the function at-3 is $\frac{2}{25}$.