# First derivative test help

1. Jun 14, 2010

### jinx007

The curve y = x^2√(1 − x^2) for x ≥ 0 and its maximum point M. Find the coordinates of M (x coordinates) ^ --> means power

dy/dx is zero at stationary point

so i make use of the chain rule

and after simplification i get my answer as

x^3 = 2 x

so x^2 = 2 and x = square root of 2

but unfortunately the answer in the marking scheme
is square root 2/3

2. Jun 14, 2010

### rl.bhat

Re: Differenciation

$$y = x^2sqrt{(1-x^2)}$$

$$\frac{dy}{dx} = x^2\frac{-2x}{2\sqrt{(1-x^2)}} + \sqrt{(1-x^2)}2x = 0$$

$$\frac{-x^3}{\sqrt{(1-x^2)}} + \sqrt{(1-x^2)}2x = 0$$

Now simplify.

3. Jun 14, 2010

### HallsofIvy

Staff Emeritus
Re: Differenciation

We can't tell you where your error is until you tell us how you got this! Your error is somewhere in that calculation.

4. Jun 15, 2010

### Susanne217

Re: Differenciation

There is a mistake cause I get

$$y = x^2\sqrt{(1-x^2)}$$

$$\frac{dy}{dx} = \frac{-3\cdot x^3-2 \cdot x}{\sqrt{(1-x)} \cdot \sqrt{(1+x)}}$$

and then use the "First derivative test" shown here

http://www.math.hmc.edu/calculus/tutorials/extrema/

to find the maximum point of the function

Last edited: Jun 15, 2010
5. Jun 15, 2010

### Mentallic

Re: Differenciation

What you have is wrong. Show us what you did and we'll steer you in the right direction. The key here is to notice that you need to use the product rule.

Last edited: Jun 15, 2010
6. Jun 15, 2010

### Susanne217

Re: Differenciation

Dude according to Wolfram integrator my result is correct..

7. Jun 15, 2010

### Mentallic

Re: Differenciation

So you didn't actually calculate the derivative yourself? Because with my understanding of derivatives, what rl.bhat got is correct and what you have is not what he has... calculator or not...

8. Jun 15, 2010

### Susanne217

Re: Differenciation

I did calculate it and get the result above !

9. Jun 15, 2010

### Mentallic

Re: Differenciation

Then post what you did so I can point at where you've gone wrong.

10. Jun 15, 2010

### Susanne217

Re: Differenciation

you got me pappa

after retrying the calculation I get

$$f'(x) = 2x\cdot \sqrt{(1-x^2)} + x^2 \cdot \frac{-2x}{2\cdot \sqrt{1-x^2}}$$

but still the first derivative test is the way to go to find the maximum of f..

11. Jun 15, 2010

### Mentallic

Re: Differenciation

There you go

12. Jun 16, 2010

### jinx007

Re: Differenciation

hey its so strange, i got the answer that is square root 2/3..

I make use of the dy/dx = v du - u dv / v^2

so i put x^2 / (1-x^2)^-1/2 and i solve...woOo so puzzling.. anyway thanks to all who helped me..!!!!

13. Jun 16, 2010

### Mentallic

Re: Differenciation

It's not puzzling at all that you apply a different rule of differentiation to solve the same function! If it were different, there are obvious flaws in our "rules".

It works for anything you like, say we wanted to differentiate x2, well we use the obvious rule to turn it into 2x, but say we instead used the quotient rule on $$\frac{1}{x^{-2}}$$ then you would find the same answer, even though more work is required

But be wary, while it may not seem like the same answer at first sight, they are still equivalent so you would need to simplify/re-arrange in order to show they're equal.

14. Jun 16, 2010

### HallsofIvy

Staff Emeritus
Re: Differenciation

You could also have done that as
$$y= x^2(1- x^2)^{1/2}$$
with the chain rule and product rule.

15. Jun 16, 2010

### Gregg

Re: Differenciation

Another way that works is

$$y^2=x^4-x^6$$

$$2yy' = 4x^3-6x^5$$

$$y' = \frac{4x^3-6x^5}{2\sqrt{x^4-x^6}}$$

of course y'=0 implies

$$4x^3-6x^5 = 0$$

$$x = \sqrt{2}/3$$ works