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anonymous
 5 years ago
How do you differentiate root(cosx) by first principles
anonymous
 5 years ago
How do you differentiate root(cosx) by first principles

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(\sqrt(\cos(x)))' = (\sqrt(y))' = y' . (y^{1\over2})' = y' . {1 \over {2.\sqrt(y)}} \] \[(\cos(x))' . {1 \over {2.\sqrt{\cos(x)}}} = {\ sin (x) \over {{2\sqrt{\cos(x)}}}}\] I guess it's right... hard to do it out the paper, lol.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(\sqrt{\cos x})'=\lim_{h\rightarrow 0}\frac{\sqrt{\cos(x+h)}\sqrt{\cos x}}{h}= \lim_{h\rightarrow 0}\frac{\cos(x+h)\cos x}{h(\sqrt{\cos(x+h)}+\sqrt{\cos x})}=\] \[=\lim_{h\rightarrow 0}\frac{\cos x\cos h\sin x\sin h\cos x}{h(\sqrt{\cos(x+h)}+\sqrt{\cos x})}=\quad,\quad\cos h=1,\sin h=h\] \[=\lim_{h\rightarrow 0}\frac{\cos xh\sin x\cos x}{h(\sqrt{\cos(x+h)}+\sqrt{\cos x})}=\lim_{h\rightarrow 0}\frac{h\sin x}{h(\sqrt{\cos(x+h)}+\sqrt{\cos x})}=\] \[=\frac{\sin x}{2\sqrt{\cos x}}\]
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