Differentiate y=(sinx)(3x^2) please help please !

6xsinx + 3x²cosx

Step-by-step explanation:

Differentiate using the product rule

Given y = f(x)(g(x) , then

= f(x)g'(x) + g(x)f'(x)

Here f(x) = sinx ⇒ f'(x) = cosx

g(x) = 3x² ⇒ g'(x) = 6x

Thus

= 6xsinx + 3x²cosx

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

what do u need to know

step-by-step explanation: