Question

H-E-L-P : HOW TO FIND DERIVATIVE SQUARE ROOT OF 8-6*X^2

Answer 1

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

Answer 2

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

Answer 3

can we do this step by step please.

Answer 4

the steps are what i wrote above
take the derivative of \(8-6x^2\) and by the power rule, you get \(-12x\)
that goes in the numerator
in the denominator you put \(2\sqrt{8-6x^2}\)

Answer 5

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

Answer 6

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

Answer 7

the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\)
therefore by the chain rule the derivative of \(\frac{1}{f}\) is \(\frac{f'}{2\sqrt{f}}\) where \(f\) is any function

Answer 8

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

Answer 9

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

Answer 10

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