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mathcalculus
Group Title
can someone help me find the derivative of a square root step by step please.
 one year ago
 one year ago
mathcalculus Group Title
can someone help me find the derivative of a square root step by step please.
 one year ago
 one year ago

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mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
Sqrt(25*x^2)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i did this: 25x^2) ^1/2
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
then 1/2(25x^2)
 one year ago

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@zepdrix
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Woops you didn't apply the power rule correctly.\[\large (25x^2)^{1/2}\]Derivative gives us,\[\large \frac{1}{2}(25x^2)^{1/21} \qquad=\qquad \frac{1}{2}(25x^2)^{1/2}\]But we also have to apply the chain rule after that, multiplying by the derivative of the inside.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
uhoh got me there... may i ask how?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i think i kind of do but im not sure how to proceed it.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
\[\large \color{royalblue}{\left[(25x^2)^{1/2}\right]'} \qquad = \qquad \frac{1}{2}(25x^2)^{1/2}\color{royalblue}{(25x^2)'}\]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,\[\large \frac{1}{2}(25x^2)^{1/2}\color{orangered}{(10x)}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Confused? :O Too much stuff going on?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im a little confused with how you got to 10x
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
It's the derivative of that blue thing. Take the derivative of each term inside the brackets, separately.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i see but what happens to the half?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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,\[\large f(x)=x^{1/2}\]\[\large g(x)=25x^2\] \[\large f(g(x))=(25x^2)^{1/2}\] Taking the derivative using the chain rule:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\] \[\large =\left[(25x^2)^{1/2}\right]'\left[25x^2\right]'\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
dw:1375768509806:dwIf 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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
It's a tricky concept to get used to :(
 one year ago

Psymon Group TitleBest ResponseYou've already chosen the best response.0
I like that one :P
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Uh oh, your brain esplode? :( Want to see a nice easy example? One which includes no potatoes? :o
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i understand the 10x but what happens to the 1/2 exponent.
 one year ago

Psymon Group TitleBest ResponseYou've already chosen the best response.0
Next time we should do f ' (walmart)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
We're making a copy of the `inside` and taking it's derivative. The 1/2 is on the `outside`.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
We're not making a copy of the `outside`, if that makes sense :o
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i know the chain rule but i dont understand the other one
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
after you find 10x then what
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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!
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
That last example was so ... unnecessary lol... I should delete that.
 one year ago

Psymon Group TitleBest ResponseYou've already chosen the best response.0
Haha, we need the humor with these things sometimes, though : )
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
noooo dont lol
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
don't delete :) thank you
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
now, the critical points are= x=10?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i mean for x
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i know how to solve for y
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Oh since addition is commutative, we can rearrange things a bit just to make it look prettier.\[\large \frac{1}{2}(25x^2)^{1/2}\color{orangered}{(10x)} \qquad=\qquad \frac{5x}{\sqrt{25x^2}}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Since `multiplication` is commutative* Now we need to find critical points?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
I kinda threw a bunch of steps in there when I simplified. Did any of that confuse you? :o
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
the chain rule, yes
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh simplied? uhhh no not really. i know how you did that.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh wait i understand th chain rule now. simply find the derivative of the "inner" which to me... is 25x^2
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
yah that's the inner thing :o take it's derivative, and multiply by it.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
rewrite and first and multiply by the inner's derivative
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay i caught up with ya.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
now to find that x.....
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
because they;re asking for the critical points.
 one year ago

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@zepdrix
 one year ago

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@Psymon
 one year ago

Psymon Group TitleBest ResponseYou've already chosen the best response.0
I didn't want to interfere xD
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
it's okay, team work =]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@Psymon
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
To find critical points we set the derivative function equal to zero. Critical points exist anywhere the derivative is zero, or does not exist.\[\large 0=\frac{5x}{\sqrt{25x^2}}\] So we can determine that a critical point exists when:\[\large 0=5x \qquad\qquad\qquad 0=\sqrt{25x^2}\]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
those are not critical points..... :/ it should be written like this.. ex: (0,0)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
From that first zero? Hmm it looks like if we divide both sides by 5, we get x=0.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
I'm not sure where you're getting 5 from :o
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
No, normally you write critical points as x= Max and min are written as coordinate pairs. Maybe your teacher does it differently though :)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oops! right x=0
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
How bout this: \[\large 0=\sqrt{25x^2}\] What critical points is it giving us?
 one year ago

Psymon Group TitleBest ResponseYou've already chosen the best response.0
Yeah, I write crit points as x = as well, only a coordinate pair for min and max. Just to kinda back up zep, lol.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
does not exist?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Squaring both sides gives us,\[\large 0=25x^2\]Solving for x gives us,\[\large x=\pm \sqrt{2/5}\]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
wait a min i had that... :/ but it was x= sqrt(2/5)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
because the  is there^..
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oops, right again.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@ zepdrix thank you so much for your time and effort. i appreciate lot's! goodnight.
 one year ago
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