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mathcalculus
 2 years ago
can someone help me find the derivative of a square root step by step please.
mathcalculus
 2 years ago
can someone help me find the derivative of a square root step by step please.

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mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0i did this: 25x^2) ^1/2

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Woops 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.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0uhoh got me there... may i ask how?

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0i think i kind of do but im not sure how to proceed it.

zepdrix
 2 years ago
Best 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,\[\large \frac{1}{2}(25x^2)^{1/2}\color{orangered}{(10x)}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Confused? :O Too much stuff going on?

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0im a little confused with how you got to 10x

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3It's the derivative of that blue thing. Take the derivative of each term inside the brackets, separately.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0i see but what happens to the half?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3That 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
 2 years ago
Best ResponseYou've already chosen the best response.3The 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
 2 years ago
Best ResponseYou've already chosen the best response.3I 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:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\] \[\large =\left[(25x^2)^{1/2}\right]'\left[25x^2\right]'\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3dw: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.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3It's a tricky concept to get used to :(

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

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

Psymon
 2 years ago
Best ResponseYou've already chosen the best response.0Next time we should do f ' (walmart)

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

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0i know the chain rule but i dont understand the other one

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0after you find 10x then what

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3We 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
 2 years ago
Best ResponseYou've already chosen the best response.3That last example was so ... unnecessary lol... I should delete that.

Psymon
 2 years ago
Best ResponseYou've already chosen the best response.0Haha, we need the humor with these things sometimes, though : )

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0don't delete :) thank you

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0now, the critical points are= x=10?

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0i know how to solve for y

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Since `multiplication` is commutative* Now we need to find critical points?

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0the chain rule, yes

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0oh simplied? uhhh no not really. i know how you did that.

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3yah that's the inner thing :o take it's derivative, and multiply by it.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0rewrite and first and multiply by the inner's derivative

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0okay i caught up with ya.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0now to find that x.....

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0because they;re asking for the critical points.

Psymon
 2 years ago
Best ResponseYou've already chosen the best response.0I didn't want to interfere xD

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0it's okay, team work =]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3To 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:\[\large 0=5x \qquad\qquad\qquad 0=\sqrt{25x^2}\]

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

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3I'm not sure where you're getting 5 from :o

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

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

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

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

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0because the  is there^..

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Hmm 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.

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0oops, right again.

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