Question

Use implicit differentiation to find the slope of the tangent line to the curve -3x^2 - 1xy -2y^3 = -54 at the point (-1, 3)
I am really frustrated with this

Answer 1

show mw what youve tried doing

Answer 2

The derivative I got was y' = -3x^2 / -x-y-6y^2

Answer 3

The derivative of -3x^2 is -6x

Answer 4

I used product rule for -1xy

Answer 5

and -xy ?

Answer 6

and i got -y + -xy

Answer 7

nope

Answer 8

ah

Answer 9

so what would it be?

Answer 10

-y-xy'

Answer 11

ok so we are at -6x - y -xy'?

Answer 12

product rule is:

-x'y - xy' ; since x' = 1 , we can simplify to

-y-xy'

Answer 13

so far yes

Answer 14

Ok and the derivative of -2y^3 is -6y^2y' , and the 54 would be trivial because when derivatives are taken it would be canceled?

Answer 15

yes and the 54 derives to 0 like all constants do

Answer 16

ok so now I would want to get the y' by itself

Answer 17

you can either plug in for x and y as is and solve for y' afterwards; or you can get y' by itself and plug in for x and y ... either way

Answer 18

Oh i see

Answer 19

Let me work this out and give you what I got?

Answer 20

sorry if i sound dumb

Answer 21

Ok are u still there?

Answer 22

I just want to make sure I have it right

Answer 23

im here

Answer 24

ok so i am at -6(-1)-3 + 1y' -6(3)^2y'

Answer 25

using the plug in method that you said

Answer 26

looks good so far except you left off the =0 :)

Answer 27

damn i am having trouble with this plugging in with the y' man i am so dumb...

Answer 28

6-3+y'-54y' = 0

3+y'(1-54) = 0

y' = 3/53

Answer 29

ok great but the 54?

Answer 30

you subtracted oh ok i see

Answer 31

:) you have to be able to retain the algebra stuff while learning the calculus stuff

Answer 32

yeah true thanks a lot man :)

Answer 33

good luck ;)

Answer 34

u saved me :)

Answer 35

and thanks