Simplification Problem before Finding Derivative

[johnstobbart]

Homework Statement

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}

Homework Equations

None, I believe.

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.

 

[Ray Vickson]

Homework Statement

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}

Homework Equations

None, I believe.

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 \frac{x}{\sqrt{x}} = \frac{1}{\sqrt{x}}? 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

 

[johnstobbart]

Why do you claim that \frac{x}{\sqrt{x}} = \frac{1}{\sqrt{x}}? 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}

 

[Millennial]

Writing your problem in terms of exponents instead of radicals make it easier:
\frac{d}{dx}\frac{x-3x^{3/2}}{x^{1/2}}
Split that up to get
\frac{d}{dx}x^{1/2}-3\frac{d}{dx}x
Then your answer is correct. To check your work, it is always good to integrate what you got:
\int \frac{1}{2\sqrt{x}}-3\,dx=\frac{1}{2}\int x^{-1/2}\,dx - 3x=\sqrt{x}-3x
which is what we started with.

 

[johnstobbart]

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.