How do you determine the Derivative with Product Rule?

Question

i need help, I don't know how to solve this problem:

y = (3x-4)squared

I need to find the derivative using the Product Rule. Can someone show me the work how to do it, I don't need the answer only.

Answer 1

Method 1
y = u v
dy/dx = u dv/dx + v du/dx

y = (3x - 4) (3x - 4)
dy/dx = 3(3x - 4) + 3(3x - 4)
dy/dx = 6(3x - 4)

Method 2
y = (3x - 4)²
let u = 3x - 4
du/dx = 3

y = u²
dy/du = 2u

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

Answer 2

You could rewrite this equation as y=(3x-4)(3x-4).

Then you could use the product rule to take the derivative: "the first term times the derivative of the second plus the second term times the derivative of the first."

So, dy/dx = (3x-4)(3) + (3x-4)(3) = 6(3x-4) = 18x-24.

Same result as the power rule

dy/dx = 2(3x-4)(3)=6(3x-4)=18x-24.

Answer 3

well
9x^2 - 24x + 16
18x - 24
with this example you could just expand it and then take the derivative.
or use the chain rule.
take the derivative of the inside
then the derivative on the outside
so
3(2)(3x-4)^(2-1)
6(3x-4)=18x - 24

Answer 4

Whether it is stupid or not doesn't matter, you can use any rule that you want to as long as it is valid.

y = (3x-4)^2

the way that this problem is written doesn't require the product rule, it requires the chain rule. So, expand it out so you can use the product rule.

y = (3x-4)(3x-4)

product rule:
y = f * g
then
y' = f'g + fg'

y' = 3 * (3x-4) + (3x-4) * 3
y' = 9x - 12 + 9x -12
y' = 18x -24

Most people would use the chain rule here:
y = (3x-4)^2
y' = 2(3x-4)*3 = 18x - 24

Answer 5

The product rule states that if you have
y = f(x) ∙ g(x) for some functions f and g, then
dy/dx = f'(x)∙g(x) + f(x)∙g'(x).

To use that in this situation, notice that
y = (3x-4)(3x-4), so that
dy/dx = (3x-4)∙d/dx (3x-4) + d/dx(3x-4)∙(3x-4)
= (3x-4)(3) + (3)(3x-4)
= 2∙3∙(3x-4)
= 6(3x-4) or 18x-24.
§

Answer 6

that would be stupid to use the product rule for this

the product rule is, given f = uv

f' = udv + vdu

but in your problem, u=v, so du=dv and the above simplifies to
f' = 2udu
which is exactly what you would have gotten with the more direct rule


u = 3x-4, thus du = 3
hence the derivative is
2(3x-4)3 =
18x-24

Answer 7

needless to say (uv)' = u^2v^2. So the respond is x^3 f(x^2). so which you desire to unravel (t-3)(6^t) = 0. t-3 is 0 while t = 3.5 6^t is 0 while t = e (evaluate the imperative and that's evident) So the respond is trivial.

Answer 8

The product rule states:

if f(x) = g(x)*h(x),
then f'(x) = g'(x)*h(x) + g(x)*h'(x)

so, in your example,

f(x) = (3x - 4)*(3x - 4). [This is in the form of g(x)*h(x)]
h(x) = (3x - 4)
g(x) = (3x - 4)

so
h'(x) = 3
g'(x) = 3

f''(x) = 3*(3x - 4) + (3x - 4)*3. [Just subsituting what I got for g(x), g'(x), h(x), and h'(x).]
f''(x) = 18x - 24

Answer 9

y = (3x -4)(3x-4)
y' = (3x -4)'(3x -4) + (3x -4)(3x -4)'
y' = 3(3x -4) + (3x -4)*3
y' = 6(3x -4)
y' = 18x - 24

Answer 10

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