Question

use the product rule to find the derivative
y=(5x^2-1)(4x+3) and how do i simplify

Answer 1

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

Answer 2

Did you know the product rule?

let (5x^2 - 1) = f(x) and (4x + 3) = g(x)

\[\large \frac{dy}{dx} = f'(x)g(x) + f(x)g'(x)\]

Answer 3

ive seen it before

Answer 4

So first...what is the derivative of

5x^2 - 1 ?

Answer 5

10x-1

Answer 6

Is it? what is the derivative of a constant?

Answer 7

i dont know

Answer 8

The derivative is the just slope of the function. What is the slope of the function \(y = 3\)?

Answer 9

The derivative of a constant is 0...so the deriviative of

5x^2 - 1

yes 10x

But the -1 gets discarded...because it is just 0

Answer 10

okay

Answer 11

It's vital that you understand why the derivative of a constant must be 0...

Answer 12

Right....if you have a function y = 3...that looks like



That right?

Well, what is the slope of that line?

Answer 13

Because as we know...the derivative ....is just the slope of a line at a given point...

Answer 14

0

Answer 15

so the next one will be 4-0

Answer 16

That is correct! the derivative of 4x + 3 will be just 4

So now...all we have when applying the product rule..is

\[\large (10x)(4x + 3) + (5x^2)(4)\]

Can you simplify that down at all?

Answer 17

no

Answer 18

how do i do that

Answer 19

60x^2+30x RIGHT?

Answer 20

@jim_thompson5910

Answer 21

Yes, that's correct, but you guys have made a mistake somewhere along the way...

Answer 22

where?

Answer 23

I'll save you the trouble of trying to spot it.
\[(10x)(4x+3) + (5x^2-1)(4)\] is what you should be simplifying

Answer 24

40x\[40x^2+30x+20x^2-4= 60x^2+30x-4\]

Answer 25

????

Answer 26

@whpalmer4

Answer 27

That is correct!

Answer 28

Thank you, i get it now.

Answer 29

I knew there was a mistake because the product of \[(5x^2−1)(4x+3)\]has an \(x^3\), an \(x^2\) term, and an \(x\) term, all of which will contribute terms to the derivative, and you didn't have the right set of terms...

Answer 30

i knew there was something wrong

Answer 31

but i didnt know what

Answer 32

As a check, you can multiply out the product and take the derivative of that:
\[(5x^2-1)(4x+3) = 20x^3+15x^2-4x-3\]\[\frac{d}{dx}[20x^3+15x^2-4x-3]=3*20x^{3-1}+2*15x^{2-1}-1*4x^{1-1}=\]\[60x^2+30x-4\checkmark\]