# Math Help - Help with one step of differentiation problem.

1. ## Help with one step of differentiation problem.

Hi, I'm working through some integral problems, and I've been able to achieve the correct solution of

$y=\sqrt{2c-x^{2}}$

I am then being asked to check my solution by differentiating it, I've also been given the solution to doing this, but I can't understand one step of the solution.

$f'\left[\left((2c-x^{2})^{\frac{1}{2}\right]=\frac{1}{2}(2c-x^{2})^{\frac{-1}{2}}(-2x)$

Where does the $(-2x)$ come from during differentiation?

Many thanks,

Shayne.

2. ## Re: Help with one step of differentiation problem.

Sorry,

the original question requires me to find the general solution of the differential equation

$\frac{dy}{dx}=-\frac{x}{y}$

for which I have the answer

$y=\sqrt{2c-x^{2}}$

It's just the one step of the checking process by differentiation I am struggling with,

Shayne

3. ## Re: Help with one step of differentiation problem.

Originally Posted by Hydralisk
Hi, I'm working through some integral problems, and I've been able to achieve the correct solution of

$y=\sqrt{2c-x^{2}}$

I am then being asked to check my solution by differentiating it, I've also been given the solution to doing this, but I can't understand one step of the solution.

$f'\left[\left((2c-x^{2})^{\frac{1}{2}\right]=\frac{1}{2}(2c-x^{2})^{\frac{-1}{2}}(-2x)$

Where does the $(-2x)$ come from during differentiation?
From the chain rule, differentiating what's inside the square root.

Many thanks,

Shayne.

4. ## Re: Help with one step of differentiation problem.

Originally Posted by Hydralisk
Hi, I'm working through some integral problems, and I've been able to achieve the correct solution of

$y=\sqrt{2c-x^{2}}$

I am then being asked to check my solution by differentiating it, I've also been given the solution to doing this, but I can't understand one step of the solution.

$f'\left[\left((2c-x^{2})^{\frac{1}{2}\right]=\frac{1}{2}(2c-x^{2})^{\frac{-1}{2}}(-2x)$

Where does the $(-2x)$ come from during differentiation?

Many thanks,

Shayne.
Confused!

Hint: (EDIT:The hint is for integration $y=\sqrt{2c-x^{2}}$, )

Use trigonometric substitution $x=\sqrt{2c}\cos\alpha$.

5. ## Re: Help with one step of differentiation problem.

I'm really sorry but both replies are confusing me.

For the chain rule, differentiating what's inside the square root does give me $-2x$, but that's not how I've been using the chain rule (so far). At the moment all I have learned is

if $y=g(u)$, where $u=f(x)$, then

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$.

I can't see this applies to the square root?

And the trigonometric substitution seems way above my level at the moment too, to be honest I'm not sure of that at all.

Sorry to be a pain in the arse,

Shayne.

6. ## Re: Help with one step of differentiation problem.

I've figured it out, using the chain rule as suggested.

By differentiating $(u)^{\frac{1}{2}$

Then differentiating $u$, where $u=(2c-x^{2})$

Then multiplying the results is

$\frac{1}{2}(2c-x^{2})^{-\frac{1}{2}}(-2x)$

Many thanks,

Shayne.

7. ## Re: Help with one step of differentiation problem.

Originally Posted by Hydralisk
I've figured it out, using the chain rule as suggested.

By differentiating $(u)^{\frac{1}{2}$

Then differentiating $u$, where $u=(2c-x^{2})$

Then multiplying the results is

$\frac{1}{2}(2c-x^{2})^{-\frac{1}{2}}(-2x)$

Many thanks,

Shayne.

Well done!