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

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

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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.
uh-oh got me there... may i ask how?
i think i kind of do but im not sure how to proceed it.
\[\large \color{royalblue}{\left[(2-5x^2)^{1/2}\right]'} \qquad = \qquad \frac{1}{2}(2-5x^2)^{-1/2}\color{royalblue}{(2-5x^2)'}\]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,\[\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)}\]
Confused? :O Too much stuff going on?
im a little confused with how you got to 10x
It's the derivative of that blue thing. Take the derivative of each term inside the brackets, separately.
isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?
i see but what happens to the half?
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:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]
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.
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:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\] \[\large =\left[(2-5x^2)^{1/2}\right]'\left[2-5x^2\right]'\]
|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.
It's a tricky concept to get used to :(
I like that one :P
Uh oh, your brain esplode? :( Want to see a nice easy example? One which includes no potatoes? :o
i understand the -10x but what happens to the -1/2 exponent.
Next time we should do f ' (wal-mart)
We're making a copy of the `inside` and taking it's derivative. The -1/2 is on the `outside`.
We're not making a copy of the `outside`, if that makes sense :o
i know the chain rule but i dont understand the other one
after you find -10x then what
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!
That last example was so ... unnecessary lol... I should delete that.
Haha, we need the humor with these things sometimes, though : )
noooo dont lol
don't delete :) thank you
now, the critical points are= x=10?
i mean for x
i know how to solve for y
Oh since addition is commutative, we can rearrange things a bit just to make it look prettier.\[\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)} \qquad=\qquad \frac{-5x}{\sqrt{2-5x^2}}\]
Since `multiplication` is commutative* Now we need to find critical points?
I kinda threw a bunch of steps in there when I simplified. Did any of that confuse you? :o
the chain rule, yes
oh simplied? uhhh no not really. i know how you did that.
oh wait i understand th chain rule now. simply find the derivative of the "inner" which to me... is 2-5x^2
yah that's the inner thing :o take it's derivative, and multiply by it.
rewrite and first and multiply by the inner's derivative
okay i caught up with ya.
now to find that x.....
because they;re asking for the critical points.
I didn't want to interfere xD
it's okay, team work =]
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:\[\large 0=-5x \qquad\qquad\qquad 0=\sqrt{2-5x^2}\]
those are not critical points..... :/ it should be written like this.. ex: (0,0)
From that first zero? Hmm it looks like if we divide both sides by -5, we get x=0.
I'm not sure where you're getting 5 from :o
No, normally you write critical points as x= Max and min are written as coordinate pairs. Maybe your teacher does it differently though :)
oops! right x=0
How bout this: \[\large 0=\sqrt{2-5x^2}\] What critical points is it giving us?
Yeah, I write crit points as x = as well, only a coordinate pair for min and max. Just to kinda back up zep, lol.
does not exist?
Squaring both sides gives us,\[\large 0=2-5x^2\]Solving for x gives us,\[\large x=\pm \sqrt{2/5}\]
wait a min i had that... :/ but it was x= sqrt(2/-5)
because the - is there^..
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.
oops, right again.
@ zepdrix thank you so much for your time and effort. i appreciate lot's! goodnight.
np c: cya

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