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Simplification Problem before Finding Derivative

 
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Aug8-12, 11:32 AM   #1
 

Simplification Problem before Finding Derivative


1. The problem statement, all variables and given/known data

I have a bit of confusion surrounding one of my basic derivative problems:

Find the derivative of the following:

\begin{equation}
f(x) = \frac{x - 3x{\sqrt{x}}}{\sqrt{x}}
\end{equation}

2. Relevant equations

None, I believe.

3. The attempt at a solution

I think I can simplify this, so I separate them:
\begin{equation}
\frac{x}{x^{1/2}} - \frac{3x. x^{1/2}}{x^{1/2}}
\end{equation}
Using the indices laws, I add and subtract:
\begin{equation}
x^{-1/2} - 3x
\end{equation}
Now it's simplified as far as I can see, so I take the derivative:
\begin{equation}
\frac{-1}{2}x^{-3/2} - 3
\end{equation}
And my final answer is:
\begin{equation}
\frac{-1}{2\sqrt{x^3}} - 3
\end{equation}
I'm confused about simplifying the exponents the way I did. If I have to use the quotient law, I'm not too sure how to apply it correctly in this case.
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Aug8-12, 11:49 AM   #2
 
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Quote by johnstobbart View Post
1. The problem statement, all variables and given/known data

I have a bit of confusion surrounding one of my basic derivative problems:

Find the derivative of the following:

\begin{equation}
f(x) = \frac{x - 3x{\sqrt{x}}}{\sqrt{x}}
\end{equation}

2. Relevant equations

None, I believe.

3. The attempt at a solution

I think I can simplify this, so I separate them:
\begin{equation}
\frac{x}{x^{1/2}} - \frac{3x. x^{1/2}}{x^{1/2}}
\end{equation}
Using the indices laws, I add and subtract:
\begin{equation}
x^{-1/2} - 3x
\end{equation}
Now it's simplified as far as I can see, so I take the derivative:
\begin{equation}
\frac{-1}{2}x^{-3/2} - 3
\end{equation}
And my final answer is:
\begin{equation}
\frac{-1}{2\sqrt{x^3}} - 3
\end{equation}
I'm confused about simplifying the exponents the way I did. If I have to use the quotient law, I'm not too sure how to apply it correctly in this case.
Why do you claim that [tex] \frac{x}{\sqrt{x}} = \frac{1}{\sqrt{x}}?[/tex] The claim is false. If, as you claim, you are just adding/subtracting exponents, then you seem to be saying that 1 - 1/2 = -1/2.

RGV
Aug8-12, 12:24 PM   #3
 
Quote by Ray Vickson View Post
Why do you claim that [tex] \frac{x}{\sqrt{x}} = \frac{1}{\sqrt{x}}?[/tex] The claim is false. If, as you claim, you are just adding/subtracting exponents, then you seem to be saying that 1 - 1/2 = -1/2.

RGV
Sorry. I made an error. Let me try fix it.
When I simplify f(x) (hopefully correctly this time), I get:
\begin{equation}
x^{1/2} - 3x
\end{equation}
Then when taking the derivative, I get:
\begin{equation}
\frac{x^{-1/2}}{2} - 3
\end{equation}
My final answer is:
\begin{equation}
\frac{1}{2\sqrt{x}} - 3
\end{equation}
Aug8-12, 01:48 PM   #4
 

Simplification Problem before Finding Derivative


Writing your problem in terms of exponents instead of radicals make it easier:
[tex]\frac{d}{dx}\frac{x-3x^{3/2}}{x^{1/2}}[/tex]
Split that up to get
[tex]\frac{d}{dx}x^{1/2}-3\frac{d}{dx}x[/tex]
Then your answer is correct. To check your work, it is always good to integrate what you got:
[tex]\int \frac{1}{2\sqrt{x}}-3\,dx=\frac{1}{2}\int x^{-1/2}\,dx - 3x=\sqrt{x}-3x[/tex]
which is what we started with.
Aug8-12, 02:35 PM   #5
 
I haven't started on integration yet, but I do know I'll be doing it soon. Thanks a lot for all your patience and time Millenial and Ray Vickson.
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