Question

Find the slope of the tangent to the curve y^3x + y^2x^2 = 6 at the point (2,1)

I know I need to use derivative, but I've never seen one arranged like this before. Help?

Answer 1

implicit ....

Answer 2

what is the derivative of y?

Answer 3

dy/dx

Answer 4

yeah, but a simpler notation is y'

Answer 5

what is the derivative of y^2 ?

Answer 6

2y?

Answer 7

and for the record, dy/dx is not correct

the derivative of y 'with respect to x' is dy/dx

otherwise its just open ended. the derivative of y is just y'

now, you are stuck in a certain mindset that the teaching method puts you in for comfort.

the derivative of y^2 is a power rule, and a chain rule ... since y is a function of something.

(y^2)' = 2y y'

Answer 8

that is the derivative of:

y^2 = 3x ? since the 'with respect to' is not given, make it open ended

2y y' = 3x'

when we know its wrt.x, x' = dx/dx = 1
when we know its wrt.y, y' = dy/dy = 1

does this make sense?

Answer 9

edit: **what is the derivative of ....

Answer 10

yes that makes sense

Answer 11

then what is the derivative of

y^3 x + y^2 x^2 = 6

[y^3 x]' + [y^2 x^2]' = [6]'

(y^3' x + y^3 x') + (y^2' x^2 + y^2 x^2') = 0

(3y^2 y' x + y^3 x') + (2y y' x^2 + y^2 2x x') = 0

solve for y', and we can assume that x' = 1

Answer 12

Alright thank you so much! It's been so long since we covered this material I forgot how to even begin!

Answer 13

youre welcome :) y' should now just be a matter of simple algebra, and you know the values of x,y to determine y'

Answer 14

it may be good to plug in x,y and then solve for y' since the rest will be numbers itll be simpler to pull it out

Answer 15

3 y' 2 + 1 + 2 y' 2^2 + 2(2) = 0

6y'+ 1 + 8y' + 4 = 0