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

Question

anonymous · Answer

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

dumbcow · Answer

you have to use product rule

dumbcow · Answer

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

jim_thompson5910 · Answer

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