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The derivative of 1/sqrt(x)

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Pithikos
#1
Jan6-11, 01:31 PM
P: 49
1. The problem statement, all variables and given/known data
Find the derivative of [tex]\frac{1}{\sqrt{x}}[/tex] using the lim definition.

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

3. 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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Dick
#2
Jan6-11, 01:38 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
Combine numerator into a single fraction. See if you get an h you can cancel with the h in the denominator.
╔(σ_σ)╝
#3
Jan6-11, 01:41 PM
╔(σ_σ)╝'s Avatar
P: 849
You could also use the definition...

[tex]f'(x)= \lim_{ x \to a} \frac{f(x)- f(a)}{x-a} [/tex].

Pithikos
#4
Jan6-11, 02:58 PM
P: 49
The derivative of 1/sqrt(x)

Thaaaank you! Problem solved! :)
Did the same to denominator and then combined the two franctions into one.
HallsofIvy
#5
Jan6-11, 04:15 PM
Math
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Thanks
PF Gold
P: 39,682
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.
SammyS
#6
Jan6-11, 04:29 PM
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PF Gold
P: 7,819
Quote Quote by Pithikos View Post
1. The problem statement, all variables and given/known data
Find the derivative of [tex]\frac{1}{\sqrt{x}}[/tex] using the lim definition.

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

3. 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]


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