Question

Find the derivative of the function at the given number.

.$f\left(x\right)=\frac{1}{\sqrt{x}},at4$

Short answer

The derivative of the function at 4is.$-\frac{1}{16}$

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=\frac{1}{\sqrt{x}},at4$

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,$a=4$ we have

$\begin{array}{c}f\text{'}\left(4\right)=\underset{h\to 0}{\lim }\frac{f(4+h)-f\left(4\right)}{h}Definitionoff\text{'}\left(4\right)\\ =\underset{h\to 0}{\lim }\frac{\left[\frac{1}{\sqrt{4+h}}\right]-\left[\frac{1}{\sqrt{4}}\right]}{h}f\left(x\right)=\frac{1}{\sqrt{x}}\\ =\underset{h\to 0}{\lim }\frac{\frac{1}{\sqrt{4+h}}-\frac{1}{2}}{h}Expand\\ =\underset{h\to 0}{\lim }\frac{2-\sqrt{4+h}}{2h\sqrt{4+h}}\times \frac{2+\sqrt{4+h}}{2+\sqrt{4+h}}Rationalizethenumerator\\ =\underset{h\to 0}{\lim }\frac{4-(4+h)}{2h\sqrt{4+h}(2+\sqrt{4+h})}Simplify\\ =\underset{h\to 0}{\lim }\frac{-h}{2h\sqrt{4+h}(2+\sqrt{4+h})}\\ =\underset{h\to 0}{\lim }\frac{-1}{2\sqrt{4+h}(2+\sqrt{4+h})}Cancelh\\ =-\frac{1}{16}\end{array}$

Therefore, the derivative of the function at4 is$-\frac{1}{16}$ .