# Thread: differentiate y = x(x+a)^1/2

1. ## differentiate y = x(x+a)^1/2

the original question was y = x squareroot x + 1

answer in the back is (-2/3, -2 squareroot 3/9)

EDIT: question is differentiate ^above^. Hence find the stationary point on the curve giving exact value.

i pretty much get no were close
sorry for not putting in the squareroot sign dont know how lol

thx

2. Originally Posted by KavX
the original question was y = x squareroot x + a

answer in the back is (-2/3, -2 squareroot 3/9

i pretty much get no were close
sorry for not putting in the squareroot sign dont know how lol

thx
use the product rule

recall that: $\displaystyle \frac d{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x)$

thus, if we have $\displaystyle y = x (x + a)^{\frac 12}$

$\displaystyle \Rightarrow y' = x' \cdot (x + a)^{\frac 12} + x \cdot \left[ (x + a)^{\frac 12 } \right]'$

now continue

3. yer i can get up to that part but i dont get what to do after and wateva i do i dont end up with the right answer the part i need help with is after that step

4. Originally Posted by KavX
yer i can get up to that part but i dont get what to do after and wateva i do i dont end up with the right answer the part i need help with is after that step
you didn't tell me what you need to do after that, you just said differentiate ...

what was the whole question?

5. woopz sorry it says differentiate ^above^. Hence find the stationary point on the curve giving exact value.

6. Originally Posted by KavX
woopz sorry it says differentiate ^above^. Hence find the stationary point on the curve giving exact value.
so the derivative is given by:

$\displaystyle y' = \sqrt{x + a} + \frac x{2 \sqrt{x + a}}$

to find the stationary points, we set the derivative to zero, thus we have:

$\displaystyle \sqrt{x + a} + \frac x{2 \sqrt{x + a}} = 0$

multiply through by $\displaystyle 2 \sqrt{x + a}$, we get:

$\displaystyle 2(x + a) + x = 0$

can you take it from here?

7. aaahhh kool thanx, i forgot to say that a = 1 sorry,

yer i can get the x value of -2/3

2(x + 1) + x = 0
2x + 2 + x = 0
3x = -2
x = -2/3

but im not sure how to get x = 2

8. Originally Posted by KavX
aaahhh kool thanx, i forgot to say that a = 1 sorry,

yer i can get the x value of -2/3

2(x + 1) + x = 0
2x + 2 + x = 0
3x = -2
x = -2/3

but im not sure how to get x = 2
who says x = 2? you said the answer was x = -2/3, which you got. to get the y-coordinate, just plug in x = -2/3 in the original expression (and plug in a = 1 as well) and solve for y

9. sorry just that another answer above of that said it had -2/3 as a x value and x = 2. my bad just read the wrong answer, anywayz thanx for all your help

10. Originally Posted by KavX