Question

Find the slope of the equation. Differentiation.

Answer 1

\[x^4-2x^2-x\]

Answer 2

Do I keep reducing the equation with differentiation?

Answer 3

You can apply power rule

Answer 4

\[\frac{ d }{ dx }x^n = nx^{n-1}\]

Answer 5

You want to find the "slope", or rather a "function that generates slope" (the "derivative of the function".)

yes, apply the power rule to each term.

Answer 6

the slope will be 24x when using differentiation rules 3 times?

Answer 7

Why are you differentiating 3 times ?

Answer 8

You only need to take the derivative once for the slope.

Answer 9

Yeah, you have to apply it to each term not 3 times

Answer 10

\[y' = 4x^3+4x-1\]

Answer 11

And you're done

Answer 12

what would be the slope? the entire equation?

Answer 13

If you want to find a slope at a point (say x=2), then plug in 2 for x.
\(\color{#000000 }{ \displaystyle f'(x)=4x^3+4x-1 }\)
\(\color{#000000 }{ \displaystyle f'(2)=4(2)^3+4(2)-1 =39}\)

Answer 14

Only have to differentiate once to find slope/tangent line

Answer 15

So, yes, the entire equation is the slope.

Answer 16

You forgot, it is -4x.

Answer 17

\(\color{#000000 }{ \displaystyle f'(x)=4x^3-4x-1 }\)

Answer 18

oh the original equation is \[x^4+2x^2-x\]

Answer 19

Oh, ok ...

Answer 20

In order to find the equation of the tangent line to the curve at the given point ( 1, 2). I would use the point-slope form?

Answer 21

Yes.

`Step  1  .`
Plug in x=1 into the derivative, to find the slope at x=1.
(Since derivative generates/is the slope of the function)

`Step  2  .`
Plug in the point (1,2) into the point slope form, with the
slope from step 1.

Answer 22

I can give you a similar example-problem as well, if you like ...

Answer 23

so slope m would be \[y'(1) = 4(1)^3+4(1)-1 = 4 + 4 - 1 = 7\]?

Answer 24

Yes, precisely!

Answer 25

Okay, I got it now. I'm used to finding slopes from equations like y = mx + b.
I can handle the rest now. :)

Answer 26

Alrighty ... good luck!