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

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mathcalculusBest ResponseYou've already chosen the best response.0
i did this: 25x^2) ^1/2
 8 months ago

zepdrixBest 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.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
uhoh got me there... may i ask how?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i think i kind of do but im not sure how to proceed it.
 8 months ago

zepdrixBest 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,\[\large \frac{1}{2}(25x^2)^{1/2}\color{orangered}{(10x)}\]
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.3
Confused? :O Too much stuff going on?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
im a little confused with how you got to 10x
 8 months ago

zepdrixBest 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.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i see but what happens to the half?
 8 months ago

zepdrixBest 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:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]
 8 months ago

zepdrixBest 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.
 8 months ago

zepdrixBest 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:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\] \[\large =\left[(25x^2)^{1/2}\right]'\left[25x^2\right]'\]
 8 months ago

zepdrixBest 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.
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.3
It's a tricky concept to get used to :(
 8 months ago

zepdrixBest 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
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i understand the 10x but what happens to the 1/2 exponent.
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Next time we should do f ' (walmart)
 8 months ago

zepdrixBest 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`.
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.3
We're not making a copy of the `outside`, if that makes sense :o
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i know the chain rule but i dont understand the other one
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
after you find 10x then what
 8 months ago

zepdrixBest 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!
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.3
That last example was so ... unnecessary lol... I should delete that.
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Haha, we need the humor with these things sometimes, though : )
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
don't delete :) thank you
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
now, the critical points are= x=10?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i know how to solve for y
 8 months ago

zepdrixBest 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.\[\large \frac{1}{2}(25x^2)^{1/2}\color{orangered}{(10x)} \qquad=\qquad \frac{5x}{\sqrt{25x^2}}\]
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.3
Since `multiplication` is commutative* Now we need to find critical points?
 8 months ago

zepdrixBest 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
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
the chain rule, yes
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
oh simplied? uhhh no not really. i know how you did that.
 8 months ago

mathcalculusBest 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
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.3
yah that's the inner thing :o take it's derivative, and multiply by it.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
rewrite and first and multiply by the inner's derivative
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay i caught up with ya.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
now to find that x.....
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
because they;re asking for the critical points.
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
I didn't want to interfere xD
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
it's okay, team work =]
 8 months ago

zepdrixBest 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:\[\large 0=5x \qquad\qquad\qquad 0=\sqrt{25x^2}\]
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
those are not critical points..... :/ it should be written like this.. ex: (0,0)
 8 months ago

zepdrixBest 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.
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.3
I'm not sure where you're getting 5 from :o
 8 months ago

zepdrixBest 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 :)
 8 months ago

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

PsymonBest 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.
 8 months ago

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

mathcalculusBest ResponseYou've already chosen the best response.0
wait a min i had that... :/ but it was x= sqrt(2/5)
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
because the  is there^..
 8 months ago

zepdrixBest 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.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
oops, right again.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@ zepdrix thank you so much for your time and effort. i appreciate lot's! goodnight.
 8 months ago
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