# Thread: finding derivative dy / dx

1. ## finding derivative dy / dx

Hi guys it's me first post here and have a strange problem with finding the derivative of :

y = (x^2 + 2x^3/2 - 5x^1/2) divided by x^1/2

The solutions ends up with x^3/2 + 2x - 5 while I end up with x^1/2 + 2x - 5

I basicaly bring up the bottom part and everything is okay except for the first term. The term x^2 to me ends up as x^2- 1/2

I can find the derivate of the solution but I can't get to the solution required to be able to move onto finding dy/dx.

2. Originally Posted by thekrown
Hi guys it's me first post here and have a strange problem with finding the derivative of :

y = (x^2 + 2x^3/2 - 5x^1/2) divided by x^1/2

The solutions ends up with x^3/2 + 2x - 5 while I end up with x^1/2 + 2x - 5

I basicaly bring up the bottom part and everything is okay except for the first term. The term x^2 to me ends up as x^2- 1/2

I can find the derivate of the solution but I can't get to the solution required to be able to move onto finding dy/dx.
I am going to assume those fractions are exponents as otherwise you would have written 2x^3/2 as x^3

$
y = \frac{x^2 + 2x^{3/2} - 5x^{1/2}}{x^{1/2}}
$

$=\frac{x^2}{x^{1/2}}+2\frac{x^{3/2}}{x^{1/2}}-5\frac{x^{1/2}}{x^{1/2}}$

$=x^{\frac{3}{2}}+2x-5$

So $\frac{dy}{dx}=\frac{3}{2}x^{\frac{1}{2}}+2$

3. The first line is good, second is good. The third line I have a problem with.

1. Take the first term, we have x^2 with x^1/2 as denom. End result is x^3/2.

2. Take second term, we have 2x^3/2 with x^1/2 as denom. End result is 2x. The way I mathed it was 2x^3/2 - 1/2 and so we get 2/2 which is 1, henc "2x^1 or 2x". Right?

3. So how come with the first term (see 1) we get 3/2 and not just 2/2?

4. Originally Posted by thekrown
The first line is good, second is good. The third line I have a problem with.

1. Take the first term, we have x^2 with x^1/2 as denom. End result is x^3/2.

2. Take second term, we have 2x^3/2 with x^1/2 as denom. End result is 2x. The way I mathed it was 2x^3/2 - 1/2 and so we get 2/2 which is 1, henc "2x^1 or 2x". Right?

3. So how come with the first term (see 1) we get 3/2 and not just 2/2?

$=\frac{x^2}{x^{1/2}}+2\frac{x^{3/2}}{x^{1/2}}-5\frac{x^{1/2}}{x^{1/2}}$

$\displaystyle {x^{2-\frac{1}{2}}+2x^{\frac{3}{2}-\frac{1}{2}}-5x^{\frac{1}{2}-\frac{1}{2}}}$

$2-\frac{1}{2}=\frac{3}{2}$ so

$=x^{\frac{3}{2}}+2x-5$

5. Oh my god how could I not see this! I have been doing much much to much math lately... my eyes are playing tricks on me!

I kept seeing it as 2/2 and not 4/2 my my... thank you. Thank you very much. I really appreciate it.