Question

find the derivative of the square root of (2x^3+4x^2+2x+1)

Answer 1

\[y=(f(x))^\frac{1}{2}=>y'=\frac{1}{2}(f(x))^{\frac{1}{2}-1}f'(x)\]

Answer 2

so it would be ( 6x^2+8x+2) ^1/2?

Answer 3

\[f(x)=2x^3+4x^2+2x+1 => f'(x)=6x^2+8x+2\]
just plug into what i wrote above

Answer 4

1/2 ( 6x^2+8x+2)

Answer 5

sorry my calc skills r rusty

Answer 6

\[y'=\frac{1}{2}(f(x))^{\frac{1}{2}-1}f'(x)\]
i gave you f(x)
and i gave you f'(x)
just plug those in