Question

How do you differentiate root(cosx) by first principles

Answer 1

\[(\sqrt(\cos(x)))' = (\sqrt(y))' = y' . (y^{1\over2})' = y' . {1 \over {2.\sqrt(y)}} \]
\[(\cos(x))' . {1 \over {2.\sqrt{\cos(x)}}} = {\ -sin (x) \over {{2\sqrt{\cos(x)}}}}\]

I guess it's right... hard to do it out the paper, lol.

Answer 2

\[(\sqrt{\cos x})'=\lim_{h\rightarrow 0}\frac{\sqrt{\cos(x+h)}-\sqrt{\cos x}}{h}=
\lim_{h\rightarrow 0}\frac{\cos(x+h)-\cos x}{h(\sqrt{\cos(x+h)}+\sqrt{\cos x})}=\]
\[=\lim_{h\rightarrow 0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h(\sqrt{\cos(x+h)}+\sqrt{\cos x})}=\quad,\quad\cos h=1,\sin h=h\]
\[=\lim_{h\rightarrow 0}\frac{\cos x-h\sin x-\cos x}{h(\sqrt{\cos(x+h)}+\sqrt{\cos x})}=\lim_{h\rightarrow 0}\frac{-h\sin x}{h(\sqrt{\cos(x+h)}+\sqrt{\cos x})}=\]
\[=\frac{-\sin x}{2\sqrt{\cos x}}\]