## mathcalculus Group Title H-E-L-P : HOW TO FIND DERIVATIVE SQUARE ROOT OF 8-6*X^2 one year ago one year ago

1. mathcalculus Group Title

$\sqrt{8-6*X^2}$

2. satellite73 Group Title

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

3. mathcalculus Group Title

can we do this step by step please.

4. satellite73 Group Title

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 $$2\sqrt{8-6x^2}$$

5. mathcalculus Group Title

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

6. satellite73 Group Title

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

7. satellite73 Group Title

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

8. satellite73 Group Title

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

9. satellite73 Group Title

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

10. mathcalculus Group Title

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