Question

derivative of y= the square root of x(3x-1)

Answer 1

\[y=\sqrt{x}(3x-1)\]

Answer 2

you have to use product rule

Answer 3

(fg)' = f'g + fg'

let
f = sqrt(x) = x^(1/2)
g = 3x-1

f' = 1/2x^(-1/2)
g' = 3

Answer 4

y = sqrt(x)*(3x-1)


y' = d/dx(sqrt(x)*(3x-1))


y' = d/dx(sqrt(x))*(3x-1) + sqrt(x)*d/dx(3x-1)


y' = d/dx(x^(1/2))*(3x-1) + sqrt(x)*d/dx(3x-1)


y' = (1/2)x^(1/2-1)*(3x-1) + sqrt(x)*(3)


y' = (1/2)x^(-1/2)*(3x-1) + 3*sqrt(x)


y' = (1/(2x^(1/2)))*(3x-1) + 3*sqrt(x)


y' = (1/(2*sqrt(x)))*(3x-1) + 3*sqrt(x)


So

\[\Large y^{\prime} = \frac{1}{2\sqrt{x}}(3x-1) + 3\sqrt{x}\]