How do you find the derivative using the product rule?

Question

Exapmle(3x-2x^2)(5+4x)

Answer 1

d/dx (uv) = ( u' )v + u( v' )

u = 3x - 2x^2
v = 5 + 4x

derivative = (3 - 4x)(5 + 4x) + (4)(3x - 2x^2)

Answer 2

The product rule says:
(d/dx)(uv) = u(dv/dx) + v(du/dx).

For your example, put u = 3x - 2x^2 and v = 5 + 4x.
Then dv/dx = 4, and du/dx = 3 - 4x.
Applying the formula, the required derivative is:
(3x - 2x^2)*4 + (5 + 4x)(3 - 4x) = 12x - 8x^2 + 15 + 12x - 20x -16x^2
= 8x^2 + 4x + 15.

Answer 3

If u = v * w

the product rule states:

u' = v' * w + v * w'

So let u = (3x - 2x^2) * (5 + 4x)

where
v = 3x - 2x^2
and
w = 5 + 4x

So:
v' = 3 - 4x
w' = 4

Therefore:
u' = v' * w + v * w'
= (3-4x) * (5+4x) + (3x -2x^2) * (4)
= 15 - 8x - 16x^2 + 12x - 8x^2
= 15 + 4x -24x^2

Answer 4

take the derivative of the first multiply by the second+ the derivative of the second multiplied by the first--

in this case

(3+4*x)*(5+4*x)+4*(3*x-2*x^2)
(3+4*x)*(5+4*x)+12x-8*x^2



is the answer

Answer 5

f (x) = (3x - 2x²) (5 + 4x)
f `(x) = (3 - 4x) (5 + 4x) + 4(3x - 2x²)
f `(x) = 15 + 12x - 20x - 16x² + 12x - 8x²
f `(x) = 15 + 4x - 24x²

Answer 6

y=(3x-2x^2)(5+4x).
dy/dx=(3-4x(5+4x)+(3x-2x^2)(+4)
=-16x^2-8x+15+12x-8x^2
=.-24x^2+4x+15.ANS.

Answer 7

u'v+v'u
derivative of first expression x second expression + derivative of second expression x first expression...