Differentiate (sqrtx)cos(sqrtx).
Chain rule and product rule... \[\Large \frac{d}{dx}\left[\sqrt{x}\cos(\sqrt x)\right]\]
As a refresher: Chain Rule: \[\Large \frac{d}{dx}{f\left(g(x)\right)}= f'\left(g(x)\right)g'(x)\] Product Rule: \[\Large \frac{d}{dx}\left[f(x)g(x)\right]= f'(x)g(x) + f(x) g'(x)\]
First apply the product rule, though...
Would it be 1/2x^(-1/2)cosx^(1/2)+x^1/2(1/2-sinx^-1/2)
Yikes... that's messy... lol, give me a sec to digest it ^_^
Mind telling me how you got the second term?
x^1/2(1/2-sinx^-1/2)
you leave x^1/2 by itself and isn't the derivative of cos(sqrtx) -sin(1/2^x^(-1/2)
really cannot understand that... could you draw it instead? There's the draw option... what's the derivative of \(\large \cos \sqrt x\)
|dw:1382923215980:dw|
I meant draw it, you know, free drawing...so that you can use symbols...
Is this what you want to say: \[\Large \sin \left(\frac1{2\sqrt x }\right)\]?
Or rather, negative of that, sorry.
\[-\sin \frac{ 1 }{ 2} \chi ^{-1/2}\]
yes that is what I am trying to say
Well then, it's wrong ^_^
Chain rule...
so what is the correct answer
Let's find out together.
\[\Large \frac{d}{dx}\cos(\sqrt x)\]
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)
\[-\sin (\sqrt{x})(-\frac{ 1 }{ 2}x ^{-1/2})\]
Still something wrong...
\[-\sin(\sqrt{x})(\frac{ 1 }{ 2}x ^{-1/2})\]
Better. \[\Large -\frac1{2\sqrt x}\sin(\sqrt x)\]
So, use that derivative when evaluating the derivative of \(\large \cos \sqrt x\) And what's your final answer?
\[\frac{ 1 }{2 \sqrt{x}}(\cos \sqrt{x})+(-\frac{ 1 }{ 2\sqrt{x} }\sin \sqrt{x})\]
You forgot something in the second term... (and yes, differentiating can be this dizzying)
did I forget to do the chain rule
Well, you forgot a certain factor...
what do you mean?
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