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\[\sqrt{8-6*X^2}\]

the derivative of \(\sqrt{f(x)}\) is
\[\frac{f'(x)}{2\sqrt{f(x)}}\]

can we do this step by step please.

oh... why do we put that as the denominator?

all that is left to do is cancel a 2 top and bottom

for example, the derivative of \(\sqrt{\sin(x)}\) is
\[\frac{\cos(x)}{2\sqrt{\sin(x)}}\]

the derivative of
\[\sqrt{3x^2+2x}\] is
\[\frac{6x+2}{2\sqrt{3x^2+2x}}\]

after i have -12x/ 2radical 8-6x^2.. how can i find the critical points?