Question

If 2 x^2+ 2 x + xy = 3 and y( 3 ) = -7, find y'( 3 ) by implicit differentiation.

Answer 1

help anyone???

Answer 2

y = (3 - 2x - 2x^2)/x
Then differentiate? I don't remember the terminology "implicit differentiation"

Answer 3

its just like finding the derivative

Answer 4

but just looking at the y term as a function f(x)

Answer 5

i think....

Answer 6

Using implicit differentiation, I suspect this has the form:\[y \prime = - \frac{2 + 4x + y}{x}\]Use the point you have to find the value for y and solve for y'.

Answer 7

how?

Answer 8

Think about it this way, let y be a function of x. Then we have f(x, y(x)). When we take the derivative of it, we take D{f(x,y)}, so we get: 4x + 2 + D{xy} = D{3} -> D{xy} = -4x - 2 Then, remember the product rule: D{xy} = y + xy', so we are left with:
y + xy' = -4x - 2 -> y' = -(4x + 2 + y)/x

Answer 9

ok i see....so i use y=3 and input this into y prime?

Answer 10

Yeah, you have the point (3,-7). Put x = 3 and y = -7 and solve for y'

Answer 11

how does y( 3 ) = -7 come into play?

Answer 12

-7/3...thanks

Answer 13

:-) No problem.

Answer 14

\[2 x^2+ 2 x + xy = 3\implies 4x+2+y+xy'=0 \]\[\implies y'=-\frac{4x+2+y}{x}~since~y(3)=-7 \implies y'(3)=-\frac{4(3)+2+(-7)}{3}=\frac{-7}{3}\]

Answer 15

Got the request a little late, hope my input is worthwhile.