Question

(a) the derivative of a function ( f ) at ( x ), denoted by ( f(x) ), is ( f(x)=lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h} ) provided that the limit exists. use the definition of the derivative of ( f ) at ( x ) to find the derivative of the given function ( f(x) ) when ( x = 4 ).

( f(4)=lim_{h
ightarrow0}\frac{left\frac{1}{4 + h}
ight-left\frac{1}{x}
ight}{h} )

evaluate the limit expression to find ( f(4) ).

( f(4)=\frac{1}{16} ) (type an integer or a fraction.)

Explanation:

Step1: Simplify the numerator

$$\begin{align*} \frac{1}{4 + h}-\frac{1}{4}&=\frac{4-(4 + h)}{4(4 + h)}\\ &=\frac{4-4 - h}{4(4 + h)}\\ &=\frac{-h}{4(4 + h)} \end{align*}$$

Step2: Substitute into the limit expression

$$\begin{align*} f^{\prime}(4)&=\lim_{h ightarrow0}\frac{\frac{-h}{4(4 + h)}}{h}\\ &=\lim_{h ightarrow0}\frac{-h}{4(4 + h)\cdot h}\\ &=\lim_{h ightarrow0}\frac{-1}{4(4 + h)} \end{align*}$$

Step3: Evaluate the limit

Substitute \(h = 0\) into \(\frac{-1}{4(4 + h)}\), we get \(\frac{-1}{4\times(4+0)}=-\frac{1}{16}\)

Answer:

\(-\frac{1}{16}\)