# Another Calculus Question?

The derivative (dy/dx) of function

y= 2x2(3-x2)5

is:

y=2x^2(3-x^2)^5

### 3 Answers

- King LeoLv 78 months agoFavourite answer
Product & Chain Rules:

y = uv

where u = u(x) and v = v(x)

dy/dx = v du/dx + u dv/dx

y = 2x² ( 3 - x² )⁵

dy/dx = ( 3 - x² )⁵ ( 4x ) + 5( 2x² )( 3 - x² )⁴( -2x )

factor out ( 3 - x² )⁴ and simplify

dy/dx = ( 3 - x² )⁴ [ ( 4x )( 3 - x² ) + 5( 2x² )( -2x ) ]

dy/dx = 12x ( 3 - x² )⁴ ( 1 - 2x² )

- Log in to reply to the answers

- ComoLv 78 months ago
Assuming that should be shown as :-

y = 2x^2 (3 - x^2)^5

dy/dx = (4x) (3 - x^2)^5 + 5 (3 - x^2)^4 (-2 x) ( 2x^2 )

dy/dx = (4x) (3 - x^2)^5 - ( 20 x^3 ) ( 3 - x^2 )^4

dy/dx = (4x) (3- x^2)^4 ( 3 - x^2 - 5x^2 )

dy/dx = (4x) ( 3 - x^2 )^4 ( 3 - 6 x^2 )

dy/dx = (12x) ( 3 - x^2 )^4 ( 1 - 2 x^2 )

- Log in to reply to the answers

- 8 months ago
Use the product rule and the chain rule

u = x^2

u' = 2x

v = (3 - x^2)^5

v' = 5 * (3 - x^2)^(4) * (-2x) = -10x * (3 - x^2)^4

y = 2 * u * v

y' = 2 * (u * v' + v * u')

y' = 2 * (x^2 * (-10x) * (3 - x^2)^(4) + 2x * (3 - x^2)^5)

y' = 2 * (3 - x^2)^(4) * (-10x * x^2 + 2x * (3 - x^2))

y' = 2 * (3 - x^2)^(4) * (-10x^3 + 6x - 2x^3)

y' = 2 * (3 - x^2)^(4) * (6x - 12x^3)

y' = 2 * 6x * (1 - 2x^2) * (3 - x^2)^(4)

y' = 12 * x * (1 - 2x^2) * (3 - x^2)^4

- Log in to reply to the answers