Question

Find g'(x) when g(x) = (2x^3 + 8x) ^ 1/2

Answer 1

Are you familiar with the chain rule?

Answer 2

Basically to derive this equation, you
take the derivative of
(2x^3 + 8x)^1/2 , with the power rule.
and
derive
2x^3 + 8x , seperately. and then multiply the two results

Answer 3

The power rule.
\[\large x^n = nx^{n-1}\]

Answer 4

Do you understand thus far?

Answer 5

no for the power rule how would I use that in this equation?

Answer 6

(2x^3+8x)^1/2
becomes
1/2(2x^3+8x)^-1/2

Answer 7

would I move the 1/2 infront?

Answer 8

oh and then 1/2(2x^3 + 8x) ^-1/2 would multiple the 2x^3 +8x

Answer 9

\[\large \frac{ (2x^3 + 8x)^{-1/2} }{ 2 } = \frac{ 1 }{ 2\sqrt{2x^3+8x} }\]

Answer 10

Now, take the derivative of 2x^3 + 8 then multiply.

Answer 11

6x^2+0 is the derivative?

Answer 12

2x^3 + 8x** I made a typo

Answer 13

6x^2+8

Answer 14

Correct.
Now you just multiply everything and simplify.
\[\large g'(x) = \frac{ 6x^2 + 8 }{ 2\sqrt{2x^3+8x} } = \frac{ 3x^2+4 }{ \sqrt{2x^3+8x} }\]

Answer 15

why is the \[2\sqrt{2x} thing on the bottom\]

Answer 16

because
\[\large x^{-n} = \frac{ 1 }{ x^n }\]
\[\large \sqrt{x} = x^{1/2}\]

Answer 17

oh thank you!

Answer 18

Glad to be of assistance.