can someone help me find the derivative of a square root step by step please.

- anonymous

can someone help me find the derivative of a square root step by step please.

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- schrodinger

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- anonymous

Sqrt(2-5*x^2)

- anonymous

i did this: 2-5x^2) ^1/2

- anonymous

then 1/2(2-5x^2)

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## More answers

- anonymous

@zepdrix

- zepdrix

Woops you didn't apply the power rule correctly.\[\large (2-5x^2)^{1/2}\]Derivative gives us,\[\large \frac{1}{2}(2-5x^2)^{1/2-1} \qquad=\qquad \frac{1}{2}(2-5x^2)^{-1/2}\]But we also have to apply the chain rule after that, multiplying by the derivative of the inside.

- anonymous

uh-oh got me there... may i ask how?

- anonymous

i think i kind of do but im not sure how to proceed it.

- zepdrix

\[\large \color{royalblue}{\left[(2-5x^2)^{1/2}\right]'} \qquad = \qquad \frac{1}{2}(2-5x^2)^{-1/2}\color{royalblue}{(2-5x^2)'}\]We took the derivative of the `outer` function, apply the power rule.
Now we have to multiply by the derivative of the inside. The prime is to show that we still need to take the derivative of that part.
Taking that derivative gives us,\[\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)}\]

- zepdrix

Confused? :O Too much stuff going on?

- anonymous

im a little confused with how you got to 10x

- zepdrix

It's the derivative of that blue thing.
Take the derivative of each term inside the brackets, separately.

- anonymous

isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?

- anonymous

i see but what happens to the half?

- zepdrix

That would be the product rule:\[\large \left[f(x) \cdot g(x)\right]' \qquad=\qquad f'(x)g(x)+f(x)g'(x)\]
We're dealing with a composition of functions, a function within a function.
Which tells us to apply the chain rule:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]

- zepdrix

The chain rule tells us...
After you've taken care of the outer function...
(In our case our outer function is "something" to the 1/2 power)
make a copy of the inside, take it's derivative, and multiply by it.

- zepdrix

I hate using function notation like this, but maybe it'll help you to understand the chain rule.
I find it more confusing, but to each his own.
So we have this,\[\large f(x)=x^{1/2}\]\[\large g(x)=2-5x^2\]
\[\large f(g(x))=(2-5x^2)^{1/2}\]
Taking the derivative using the chain rule:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]
\[\large =\left[(2-5x^2)^{1/2}\right]'\left[2-5x^2\right]'\]

- zepdrix

|dw:1375768509806:dw|If we have a function of a potato...
The chain rule tells us that after we take the derivative of the function, we make a copy of the potato and take it's derivative as well.

- zepdrix

It's a tricky concept to get used to :(

- Psymon

Haha

- Psymon

I like that one :P

- zepdrix

Uh oh, your brain esplode? :(
Want to see a nice easy example? One which includes no potatoes? :o

- anonymous

i understand the -10x but what happens to the -1/2 exponent.

- anonymous

lol

- Psymon

Next time we should do f ' (wal-mart)

- zepdrix

We're making a copy of the `inside` and taking it's derivative. The -1/2 is on the `outside`.

- zepdrix

We're not making a copy of the `outside`, if that makes sense :o

- anonymous

i know the chain rule but i dont understand the other one

- anonymous

after you find -10x then what

- zepdrix

We have this delicious Chicken breast seasoned with rosemary and all this good stuff. Then we stuff it with cheese and maybe even potatoes, cause that's how we roll.
When we take the derivative of this chicken, the chain rule tells us to make a copy of the cheese and potatoes. This copy will not include any of the rosemary or seasoning from the initial chicken.
After the -10x?
We're done!

- zepdrix

That last example was so ... unnecessary lol...
I should delete that.

- Psymon

Haha, we need the humor with these things sometimes, though : )

- anonymous

noooo dont lol

- anonymous

don't delete :) thank you

- anonymous

now, the critical points are= x=10?

- anonymous

i mean for x

- anonymous

i know how to solve for y

- zepdrix

Oh since addition is commutative, we can rearrange things a bit just to make it look prettier.\[\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)} \qquad=\qquad \frac{-5x}{\sqrt{2-5x^2}}\]

- zepdrix

Since `multiplication` is commutative*
Now we need to find critical points?

- zepdrix

I kinda threw a bunch of steps in there when I simplified.
Did any of that confuse you? :o

- anonymous

the chain rule, yes

- anonymous

oh simplied? uhhh no not really. i know how you did that.

- anonymous

@zepdrix

- anonymous

oh wait i understand th chain rule now. simply find the derivative of the "inner" which to me... is 2-5x^2

- zepdrix

yah that's the inner thing :o
take it's derivative, and multiply by it.

- anonymous

rewrite and first and multiply by the inner's derivative

- anonymous

okay i caught up with ya.

- anonymous

now to find that x.....

- anonymous

because they;re asking for the critical points.

- anonymous

@zepdrix

- anonymous

@Psymon

- Psymon

I didn't want to interfere xD

- anonymous

it's okay, team work =]

- anonymous

@Psymon

- zepdrix

To find critical points we set the derivative function equal to zero.
Critical points exist anywhere the derivative is zero, or does not exist.\[\large 0=\frac{-5x}{\sqrt{2-5x^2}}\]
So we can determine that a critical point exists when:\[\large 0=-5x \qquad\qquad\qquad 0=\sqrt{2-5x^2}\]

- anonymous

x=5?

- anonymous

those are not critical points..... :/ it should be written like this.. ex: (0,0)

- zepdrix

From that first zero? Hmm it looks like if we divide both sides by -5, we get x=0.

- zepdrix

I'm not sure where you're getting 5 from :o

- zepdrix

No, normally you write critical points as x=
Max and min are written as coordinate pairs.
Maybe your teacher does it differently though :)

- anonymous

oops! right x=0

- zepdrix

How bout this:
\[\large 0=\sqrt{2-5x^2}\]
What critical points is it giving us?

- Psymon

Yeah, I write crit points as x = as well, only a coordinate pair for min and max. Just to kinda back up zep, lol.

- anonymous

does not exist?

- zepdrix

Squaring both sides gives us,\[\large 0=2-5x^2\]Solving for x gives us,\[\large x=\pm \sqrt{2/5}\]

- anonymous

wait a min i had that... :/ but it was x= sqrt(2/-5)

- anonymous

because the - is there^..

- zepdrix

Hmm you missed a negative somewhere :o
Subtracting 2 from each side gives us -2 on the left.
The dividing by -5 gives us positive 2/5 on the left.

- anonymous

oops, right again.

- anonymous

@ zepdrix thank you so much for your time and effort. i appreciate lot's! goodnight.

- zepdrix

np c: cya

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