Question

The value of the derivative of
y=(x-1)(3x+2x^2)
at x=1 is....

Answer 1

Use product rule to find y'

Answer 2

Do u know product rule.

Answer 3

y = uv
y' = u'v + uv'

Answer 4

DOES IT HELP? @JosephLing

Answer 5

still not understand

Answer 6

Let u=x-1
AND v=(3x+2x^2)

Answer 7

can u find u' and v'

Answer 8

\[y'(x)=(3x+2x^2)+(x-1)(3+4x)=3x+2x^2+3x+4x^2-3-4x=6x^2+2x-3\\y'(1)=6(1)^2+2(1)-3=5\]

Answer 9

just expand it then differentiate
y = 2x^3 + + x^2 - 3x

the use the basic differentiate rule

Answer 10

thx...everyone