anonymous
  • anonymous
can someone help me find the derivative of a square root step by step please.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Sqrt(2-5*x^2)
anonymous
  • anonymous
i did this: 2-5x^2) ^1/2
anonymous
  • anonymous
then 1/2(2-5x^2)

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

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