The derivative of 1/sqrt(x)

  • Thread starter Pithikos
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    Derivative
In summary: \displaystyle = \frac{-1}{x\cdot x}\cdot \lim_{h\to 0} \ \ \frac{1}{\left(1+\frac{h}{x}\right)} \cdot \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}} } \displaystyle = \frac{-1}{x^2}\cdot \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}} \displaystyle = \frac{-1+1}{x^{2}\sqrt{x}}} \displaystyle = \frac{0}{x^{2}\sqrt{x}
  • #1
Pithikos
55
1

Homework Statement


Find the derivative of [tex]\frac{1}{\sqrt{x}}[/tex] using the lim definition.

Homework Equations


f(x)'=[tex]\frac{f(x+h)-f(x)}{h}[/tex]

The Attempt at a Solution


Keep in mind that everything bellow is for the lim as h approaches 0.

[tex]\frac{1}{\sqrt{x}}[/tex]

[tex]\Downarrow[/tex]

[tex]
\frac{
\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}
}
{h}
[/tex]

[tex]\Downarrow[/tex]

(I multiply both nominator and denominator with conjugate)

[tex]
\frac
{
\frac{1}{x+h}-\frac{1}{x}
}
{
\frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}}
}
[/tex]

After this I am totally lost..
 
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  • #2
Combine numerator into a single fraction. See if you get an h you can cancel with the h in the denominator.
 
  • #3
You could also use the definition...

[tex]f'(x)= \lim_{ x \to a} \frac{f(x)- f(a)}{x-a} [/tex].
 
  • #4
Thaaaank you! Problem solved! :)
Did the same to denominator and then combined the two franctions into one.
 
  • #5
Glad you got it solved. As a check, remember that you can write [itex]\frac{1}{\sqrt{x}}[/itex] as [itex]x^{-1/2}[/itex] and use the power rule.
 
  • #6
Pithikos said:

Homework Statement


Find the derivative of [tex]\frac{1}{\sqrt{x}}[/tex] using the lim definition.

Homework Equations


f(x)'=[tex]\frac{f(x+h)-f(x)}{h}[/tex]

The Attempt at a Solution


Keep in mind that everything bellow is for the lim as h approaches 0.

[tex]\frac{1}{\sqrt{x}}[/tex]

[tex]\Downarrow[/tex]

[tex]
\frac{
\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}
}
{h}
[/tex]

[tex]\Downarrow[/tex]

(I multiply both nominator and denominator with conjugate)

[tex]
\frac
{
\frac{1}{x+h}-\frac{1}{x}
}
{
\frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}}
}
[/tex]

After this I am totally lost..
This is perfectly fine - up to this point.
Continuing on:

[tex]\displaystyle =\frac
{ \displaystyle \frac{x-(x+h)}{(x+h)x}
}
{ \displaystyle \frac{h}{\ \sqrt{x+h}}+\frac{h}{\sqrt{x}\ \ }
} [/tex]

[tex]\displaystyle =\frac
{ \displaystyle \frac{-h}{(x+h)(x)}\ \cdot\ \displaystyle \frac{1}{h}
}
{ \displaystyle \left(\frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}}\right) \ \cdot\ \displaystyle \frac{1}{h}}
}
[/tex]

[tex]\displaystyle =\frac
{ \displaystyle \frac{-1}{(x+h)(x)}
}
{ \displaystyle \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}
}
[/tex]

Then,
[tex]\displaystyle f'(x)= \lim_{h\to 0} \ \
\frac
{ \displaystyle \frac{-1}{(x+h)(x)}
}
{ \displaystyle \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}
}
[/tex]
 

1. What is the derivative of 1/sqrt(x)?

The derivative of 1/sqrt(x) is equal to -1/2x^(-3/2).

2. How do you find the derivative of 1/sqrt(x)?

To find the derivative of 1/sqrt(x), you can use the power rule for derivatives, which states that the derivative of x^n is equal to nx^(n-1). In this case, n = -1/2, so the derivative is -1/2x^(-1/2-1) = -1/2x^(-3/2).

3. Why is the derivative of 1/sqrt(x) equal to -1/2x^(-3/2)?

The derivative of a function represents the rate of change of that function. In this case, 1/sqrt(x) can be rewritten as x^(-1/2), so the derivative is -1/2x^(-1/2-1) = -1/2x^(-3/2). This means that for every unit increase in x, the function decreases by -1/2x^(-3/2) units.

4. Can the derivative of 1/sqrt(x) be simplified?

Yes, the derivative of 1/sqrt(x) can be simplified to -1/2x^(-3/2) = -1/(2sqrt(x)).

5. What is the domain of the derivative of 1/sqrt(x)?

The domain of the derivative of 1/sqrt(x) is all real numbers except 0, since the original function 1/sqrt(x) is undefined at x = 0 and the derivative is equal to -1/2x^(-3/2).

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