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Chapter 2: Problem 18

Find the derivative by the limit process. \(f(x)=\frac{4}{\sqrt{x}}\)

Short Answer

Expert verified
The derivative of the function \(f(x)=\frac{4}{\sqrt{x}}\) by the limit process is \(f'(x) = -\frac{2}{\sqrt{x^{3}}}\).

Step by step solution

01

Rewrite the Function

First, rewrite the function \(f(x)=\frac{4}{\sqrt{x}}\) using negative exponents in order to simplify the derivation process. This yields \(f(x)=4x^{-\frac{1}{2}}\).
02

Apply the Definition of the Derivative

The derivative of a function f at a point x is defined as the limit of the difference quotient. Use this definition to rewrite the function as follows, \(f'(x) = \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\).
03

Substitute the function

Substitute the function into the definition of the derivative to get: \(f'(x) = \lim_{h \rightarrow 0}\frac{4(x+h)^{-\frac{1}{2}} - 4x^{-\frac{1}{2}}}{h}\).
04

Simplify

Perform algebraic simplifications to continue the limit process, by finding a common denominator inside the limit. This yields \(f'(x) = \lim_{h \rightarrow 0}\frac{4}{\sqrt{x+h}} - \frac{4}{\sqrt{x}} = \lim_{h \rightarrow 0} \frac{4\sqrt{x} - 4\sqrt{x+h}}{x(x+h)}\).
05

Final Calculation of the Limit

Next, evaluate the limit as h approaches 0, this yields \(f'(x) = -\frac{2}{\sqrt{x^{3}}}\). This is the derivative of the function.

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Most popular questions from this chapter

Existence of an Inverse Determine the values of \(k\) such that the function \(f(x)=k x+\sin x\) has an inverse function.

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Determine the point(s) at which the graph of \(f(x)=\frac{x}{\sqrt{2 x-1}}\) has a horizontal tangent line.
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