(a) the derivative of a function f at x, denoted by f(x), is f(x)=lim h→0 f(x + h)-f(x)/h provided that the…

(a) the derivative of a function f at x, denoted by f(x), is f(x)=lim h→0 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→0 1/(4 + h)-1/4/h (b) f(4)= -1/16 (type an integer or a fraction.) (b) f(4)= 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→0 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→0 1/(4 + h)-1/4/h (b) f(4)= -1/16 (type an integer or a fraction.) (b) f(4)= 1/16 (type an integer or a fraction.)

Answer

Explanation:

Step1: Simplify the expression

$$ \begin{align*} f^{\prime}(4)&=\lim_{h \to 0}\frac{\frac{1}{4 + h}-\frac{1}{4}}{h}\ &=\lim_{h \to 0}\frac{\frac{4-(4 + h)}{4(4 + h)}}{h}\ &=\lim_{h \to 0}\frac{\frac{4-4 - h}{4(4 + h)}}{h}\ &=\lim_{h \to 0}\frac{\frac{-h}{4(4 + h)}}{h} \end{align*} $$

Step2: Cancel out the common factor

$$ \begin{align*} f^{\prime}(4)&=\lim_{h \to 0}\frac{-h}{4(4 + h)\times h}\ &=\lim_{h \to 0}\frac{-1}{4(4 + h)} \end{align*} $$

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:

(f^{\prime}(4)=-\frac{1}{16})