Certainly! Let's evaluate the limit [tex]\(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)[/tex] for the given function [tex]\(f(x) = \frac{1}{\sqrt{x+1}}\)[/tex].
### Step-by-Step Solution:
1. State the Problem:
We need to find the limit:
[tex]\[
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\][/tex]
where [tex]\(f(x) = \frac{1}{\sqrt{x+1}}\)[/tex].
2. Write the Difference Quotient:
Start with the difference quotient:
[tex]\[
\frac{f(x+h) - f(x)}{h}
\][/tex]
3. Substitute [tex]\(f(x)\)[/tex] and [tex]\(f(x+h)\)[/tex]:
[tex]\(f(x) = \frac{1}{\sqrt{x+1}}\)[/tex], and [tex]\(f(x+h) = \frac{1}{\sqrt{x+h+1}}\)[/tex].
[tex]\[
\frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{\sqrt{x+h+1}} - \frac{1}{\sqrt{x+1}}}{h}
\][/tex]
4. Combine the Fractions:
To simplify, we need a common denominator for [tex]\(\frac{1}{\sqrt{x+h+1}} - \frac{1}{\sqrt{x+1}}\)[/tex]. The common denominator is [tex]\(\sqrt{x+h+1} \cdot \sqrt{x+1}\)[/tex]:
[tex]\[
\frac{\frac{1}{\sqrt{x+h+1}} - \frac{1}{\sqrt{x+1}}}{h} = \frac{\sqrt{x+1} - \sqrt{x+h+1}}{h \cdot \sqrt{x+h+1} \cdot \sqrt{x+1}}
\][/tex]
5. Rationalize the Numerator:
Multiply the numerator and denominator by the conjugate of the numerator [tex]\(\sqrt{x+1} + \sqrt{x+h+1}\)[/tex] to simplify:
[tex]\[
\frac{(\sqrt{x+1} - \sqrt{x+h+1}) \cdot (\sqrt{x+1} + \sqrt{x+h+1})}{h \cdot \sqrt{x+h+1} \cdot \sqrt{x+1} \cdot (\sqrt{x+1} + \sqrt{x+h+1})}
\][/tex]
6. Simplify the Expression:
The numerator becomes:
[tex]\[
(\sqrt{x+1})^2 - (\sqrt{x+h+1})^2 = (x+1) - (x+h+1) = -h
\][/tex]
So, the expression is:
[tex]\[
\frac{-h}{h \cdot \sqrt{x+1} \cdot \sqrt{x+h+1} \cdot (\sqrt{x+1} + \sqrt{x+h+1})}
\][/tex]
7. Cancel [tex]\(h\)[/tex] from Numerator and Denominator:
Canceling [tex]\(h\)[/tex] we get:
[tex]\[
\frac{-1}{\sqrt{x+1} \cdot \sqrt{x+h+1} \cdot (\sqrt{x+1} + \sqrt{x+h+1})}
\][/tex]
8. Take the Limit as [tex]\(h \to 0\)[/tex]:
As [tex]\(h \to 0\)[/tex], [tex]\(\sqrt{x+h+1} \to \sqrt{x+1}\)[/tex]:
[tex]\[
\lim_{h \to 0} \frac{-1}{\sqrt{x+1} \cdot \sqrt{x+h+1} \cdot (\sqrt{x+1} + \sqrt{x+h+1})} = \frac{-1}{\sqrt{x+1} \cdot \sqrt{x+1} \cdot (2\sqrt{x+1})}
\][/tex]
Simplify:
[tex]\[
\frac{-1}{\sqrt{x+1} \cdot \sqrt{x+1} \cdot 2\sqrt{x+1}} = \frac{-1}{2(x+1)^{3/2}}
\][/tex]
9. Conclusion:
Therefore, the limit is:
[tex]\[
\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = -\frac{1}{2(x+1)^{3/2}}
\][/tex]
The final result of the derivative of the function [tex]\(f(x)\)[/tex] is:
[tex]\[
f'(x) = -\frac{1}{2(x+1)^{3/2}}
\][/tex]
