Question

i need to find the equation of a tangent line to a given curve at a given value of x

Answer 1

y-y1 = m(x-x1) where (x1,y1) is a point on the curve. Then take the derivative of the curve, plug in x1 to the derivative, and that is your slope (or m)

Answer 2

yes , this part i know. Not sure if im doing it right though. looking for help walking through it.

Answer 3

What's the equation and x-value given?

Answer 4

y=x(cubed) - 25x + 5; x = 5

Answer 5

Okay, so f(5) = 5^3 - 25*5 + 5 = 5, so the point is (5,5)

Answer 6

i think my equation (solution) is y = 50x - 245

Answer 7

yes i got that too, so im feeling better

Answer 8

Then taking the derivative of f(x), we get f'(x) to be 3x^2 - 25, and plugging in x=5 for that, we get 50.

Answer 9

So the equation of the tangent line is y-5 = 50(x-5), or y = 50x - 245, so you are correct

Answer 10

boom!

Answer 11

thanks so much

Answer 12

No problem buddy

Answer 13

are you any good with rates of change?

Answer 14

I'm good with any calculus; what's the question?

Answer 15

D(p) = -3p(squared) + 7p + 7, where p is price, find D'(5)

Answer 16

totally lost on this stuff

Answer 17

D(p) = -3p^2 + 7p + 7, so the derivative would be D'(p) = -6p + 7. Plugging in p=5 would give us D'(5) = -23

Answer 18

okay i see what you did, why do we just throw out the 7p ?

Answer 19

I didn't, I took the derivative of 7p, which gave me 7. I got rid of the 7 because when you take the derivative of a constant (any number without a variable), then it becomes 0.

Answer 20

oh ok i get it. i seem to remember going over that in class. our professor is way behind and he's flying through the book at a rate that we can't keep up with.

Answer 21

you going to be on here awhile?

Answer 22

Yeah I'll take any other questions you have.

Answer 23

I can explain whatever you need me to explain.

Answer 24

A simple way of doing the derivative is f(x) =a*x^n , f'(x) = n*a*x^(n-1). This only holds for simple examples and cannot be used to derive something like (x^2 + 2)^2,notice the exponent outside the brackets, which requires the chain rule.

Answer 25

i want to try to bang out more of this homework in a way that I actually learn it so let me try a few more and get back to you.

Answer 26

So for 7p^1, to the power of one is implied, using the formula, a = 7, n = 1, the derivative is 7*1*p^(1-0) = 7*1*p^0 = 7*1*1 = 7

Answer 27

ya carson thats all giberrish to me. i wish looking at this stuff made sense but it just makes my eyes cross and everything goes blurry.

Answer 28

i do remember going over that though and its coming back to me now. i just need to keep pressing on through these problems so that something will stick

Answer 29

you guys are awesome. i'll be back in a few im sure. want to try a few more on my own.

Answer 30

okay, using the product rule to find the derivative where \[f(x)=(x^{2}-3x+2)(2x ^{3}-x ^{2}+4)\], is the solution \[10x ^{4}-24x ^{3}+21x ^{2}+4x-12\]

Answer 31

-24x^3 should be -28x^3, but other than that, everything else is correct.

Answer 32

what did i not multiply correctly to acheive that answer? im doing that in my head mostly, like john nash.

Answer 33

Probably multiplied wrong somewhere, with problems like this, it's really easy to miss something or add wrong.

Answer 34

okay can you walk me through this, i dont have an example in my book for finding marginal demand. D(p) =(6p+250) / (8p+19), what is the marginal demand D'(p)