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Mathematics 19 Online
OpenStudy (mony01):

Di fferentiate (sqrtx)cos(sqrtx).

terenzreignz (terenzreignz):

Chain rule and product rule... \[\Large \frac{d}{dx}\left[\sqrt{x}\cos(\sqrt x)\right]\]

terenzreignz (terenzreignz):

As a refresher: Chain Rule: \[\Large \frac{d}{dx}{f\left(g(x)\right)}= f'\left(g(x)\right)g'(x)\] Product Rule: \[\Large \frac{d}{dx}\left[f(x)g(x)\right]= f'(x)g(x) + f(x) g'(x)\]

terenzreignz (terenzreignz):

First apply the product rule, though...

OpenStudy (mony01):

Would it be 1/2x^(-1/2)cosx^(1/2)+x^1/2(1/2-sinx^-1/2)

terenzreignz (terenzreignz):

Yikes... that's messy... lol, give me a sec to digest it ^_^

terenzreignz (terenzreignz):

Mind telling me how you got the second term?

terenzreignz (terenzreignz):

x^1/2(1/2-sinx^-1/2)

OpenStudy (mony01):

you leave x^1/2 by itself and isn't the derivative of cos(sqrtx) -sin(1/2^x^(-1/2)

terenzreignz (terenzreignz):

really cannot understand that... could you draw it instead? There's the draw option... what's the derivative of \(\large \cos \sqrt x\)

OpenStudy (mony01):

|dw:1382923215980:dw|

terenzreignz (terenzreignz):

I meant draw it, you know, free drawing...so that you can use symbols...

terenzreignz (terenzreignz):

Is this what you want to say: \[\Large \sin \left(\frac1{2\sqrt x }\right)\]?

terenzreignz (terenzreignz):

Or rather, negative of that, sorry.

OpenStudy (mony01):

\[-\sin \frac{ 1 }{ 2} \chi ^{-1/2}\]

OpenStudy (mony01):

yes that is what I am trying to say

terenzreignz (terenzreignz):

Well then, it's wrong ^_^

terenzreignz (terenzreignz):

Chain rule...

OpenStudy (mony01):

so what is the correct answer

terenzreignz (terenzreignz):

Let's find out together.

terenzreignz (terenzreignz):

\[\Large \frac{d}{dx}\cos(\sqrt x)\]

terenzreignz (terenzreignz):

Again, chain rule \[\Large \frac{d}{dx}{f\left(g(x)\right)}= f'\left(g(x)\right)g'(x)\] Take f to be cos(x) and g to be sqrt(x)

OpenStudy (mony01):

\[-\sin (\sqrt{x})(-\frac{ 1 }{ 2}x ^{-1/2})\]

terenzreignz (terenzreignz):

Still something wrong...

OpenStudy (mony01):

\[-\sin(\sqrt{x})(\frac{ 1 }{ 2}x ^{-1/2})\]

terenzreignz (terenzreignz):

Better. \[\Large -\frac1{2\sqrt x}\sin(\sqrt x)\]

terenzreignz (terenzreignz):

So, use that derivative when evaluating the derivative of \(\large \cos \sqrt x\) And what's your final answer?

OpenStudy (mony01):

\[\frac{ 1 }{2 \sqrt{x}}(\cos \sqrt{x})+(-\frac{ 1 }{ 2\sqrt{x} }\sin \sqrt{x})\]

terenzreignz (terenzreignz):

You forgot something in the second term... (and yes, differentiating can be this dizzying)

OpenStudy (mony01):

did I forget to do the chain rule

terenzreignz (terenzreignz):

Well, you forgot a certain factor...

OpenStudy (mony01):

what do you mean?

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