(a) the derivative of a function f at x, denoted by f(x), is f(x) = \\lim_{h \\to 0} \\frac{f(x + h)…

(a) the derivative of a function f at x, denoted by f(x), is f(x) = \\lim_{h \\to 0} \\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\nf(4) = \\lim_{h \\to 0} \\frac{\\frac{1}{4 + h} - \\frac{1}{x}}{h}\n\nevaluate the limit expression to find f(4).\n\nf(4) = \\square (type an integer or a fraction.)

(a) the derivative of a function f at x, denoted by f(x), is f(x) = \\lim_{h \\to 0} \\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\nf(4) = \\lim_{h \\to 0} \\frac{\\frac{1}{4 + h} - \\frac{1}{x}}{h}\n\nevaluate the limit expression to find f(4).\n\nf(4) = \\square (type an integer or a fraction.)

Answer

Explanation:

Step1: Simplify the numerator

First, simplify (\frac{1}{4 + h}-\frac{1}{4}). [ \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*} ] So, (f^{\prime}(4)=\lim_{h\rightarrow0}\frac{\frac{-h}{4(4 + h)}}{h}).

Step2: Simplify the fraction

Simplify (\frac{\frac{-h}{4(4 + h)}}{h}). [ \begin{align*} \frac{\frac{-h}{4(4 + h)}}{h}&=\frac{-h}{4(4 + h)}\cdot\frac{1}{h}\ &=\frac{-1}{4(4 + h)} \end{align*} ]

Step3: Evaluate the limit

Now, find (\lim_{h\rightarrow0}\frac{-1}{4(4 + h)}). Substitute (h = 0) into (\frac{-1}{4(4 + h)}), we get (\frac{-1}{4\times(4+0)}=-\frac{1}{16}).

Answer:

(-\frac{1}{16})