Troubleshooting Differentiation of x over Root x

[batterypack]

I tried to use tex code but it seems to be caching the image on preview, so anyway

dy/dx ( x / root x) = 1 / ( 1/2 x^(-1/2) ) however the answer given to me is just ( 1/2 x^(-1/2).

What's gone wrong? Spent an hour on this and now unsure if my algebra is wrong.

And the tex code that I can't preview:

\frac{dy}{dx} \frac{x}{\sqrt{x}} =
\frac{1}{ \frac{1}{2}x^{ \frac{-1}{2} }

 

[Mark44]

I tried to use tex code but it seems to be caching the image on preview, so anyway

dy/dx ( x / root x) = 1 / ( 1/2 x^(-1/2) ) however the answer given to me is just ( 1/2 x^(-1/2).

First off, you're not taking "dy/dx" of something. dy/dx is the derivative. What you want is d/dx(x/sqrt(x)).

What's gone wrong? Spent an hour on this and now unsure if my algebra is wrong.

What went wrong is that you attempted to take the derivative of a quotient, but you didn't use the quotient rule.

\frac{d}{dx} \frac{f(x)}{g(x)} \neq \frac{f'(x)}{g'(x)}

Since x/sqrt(x) = sqrt(x) = x1/2, d/dx(x/sqrt(x)) = (1/2)x^-1/2
If you use the quotient rule, you get
\frac{x^{1/2} \cdot 1 - x \cdot (1/2)x^{-1/2}}{x}

This simplifies to (1/2)x^-1/2

And the tex code that I can't preview:

\frac{dy}{dx} \frac{x}{\sqrt{x}} =
\frac{1}{ \frac{1}{2}x^{ \frac{-1}{2} }

 

[Mark44]

I tried to use tex code but it seems to be caching the image on preview, so anyway

Yes, this is a known problem on this site.

dy/dx ( x / root x) = 1 / ( 1/2 x^(-1/2) ) however the answer given to me is just ( 1/2 x^(-1/2).

What's gone wrong? Spent an hour on this and now unsure if my algebra is wrong.

And the tex code that I can't preview:

\frac{dy}{dx} \frac{x}{\sqrt{x}} =
\frac{1}{ \frac{1}{2}x^{ \frac{-1}{2} }

 

[batterypack]

got it, thanks.