(a) the derivative of a function ( f ) at ( x ), denoted by ( f(x) ), is ( f(x)=lim_{h\rightarrow0}\frac{f(x…

(a) the derivative of a function ( f ) at ( x ), denoted by ( f(x) ), is ( f(x)=lim_{h\rightarrow0}\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 ).\n\n( f(4)=lim_{h\rightarrow0}\frac{left\frac{1}{4 + h}\right-left\frac{1}{x}\right}{h} )\n\nevaluate the limit expression to find ( f(4) ).\n\n( f(4)=\frac{1}{16} ) (type an integer or a fraction.)

(a) the derivative of a function ( f ) at ( x ), denoted by ( f(x) ), is ( f(x)=lim_{h\rightarrow0}\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 ).\n\n( f(4)=lim_{h\rightarrow0}\frac{left\frac{1}{4 + h}\right-left\frac{1}{x}\right}{h} )\n\nevaluate the limit expression to find ( f(4) ).\n\n( f(4)=\frac{1}{16} ) (type an integer or a fraction.)

Answer

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

$$ \begin{align*} f^{\prime}(4)&=\lim_{h\rightarrow0}\frac{\frac{-h}{4(4 + h)}}{h}\ &=\lim_{h\rightarrow0}\frac{-h}{4(4 + h)\cdot h}\ &=\lim_{h\rightarrow0}\frac{-1}{4(4 + h)} \end{align*} $$

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})