Question

Find the value of the slope of the tangent line to the curve x^2+2xy+3y^2=4 at the point (2,3). Using the Product Rule

Answer 1

did you need help with the derivative part?

Answer 2

I need help with all of it lol I'd like a solution and then an explanation as to how you got there

Answer 3

x^2+2xy+3y^2=4

2x + 2( 1*y + x* y ' ) + 3* 2y * y ' = 0

Answer 4

would it not be \[2x + (2x)(1. \frac{ dy }{ dx }) + (y)(2) + 6y (\frac{ dy }{ dx }) ? \]

Answer 5

and the 4 becomes a 0

Answer 6

???

Answer 7

i factored out the 2 before doing product rule on x*y

Answer 8

2 ( x * y )

Answer 9

now use the product rule on x * y

Answer 10

I don't know how

Answer 11

show me and then explain

Answer 12

(u * v) ' = u ' * v + u * v '

Answer 13

sorry I don't know how to do it like that. we write it like \[u \frac{ dv }{ dx } + v \frac{ du }{ dx }\]

Answer 14

Could you do it like that? The other way makes no sense to me

Answer 15

Sorry this is really important. I need help