Question

differentiate 6x^2 + 3xy + 2y^2 + 17y -6 = 0

Answer 1

12x + 3(dy/dx) + 4y(dy/dx) + dy/dx = 0

Answer 2

but I don't think this is right.

Answer 3

\[12x+3\left( x \frac{ dy }{ dx }+1*y \right)+4y \frac{ dy }{ dx }+17\frac{ dy }{ dx }=0\]
find \[\frac{ dy }{ dx }\]

Answer 4

Implicit differentiation, and the chain rule, product rule
\[\frac{ d }{ dx }[y] = \frac{ d }{ dy }[y]\frac{ dy }{ dx }\]

Answer 5

messed up on the 1 step, and also forgot that dy/dx of 17y is 17(dy/dx), answer is over here http://www.wolframalpha.com/input/?i=implicit+differentiation+6x%5E2+%2B+3xy+%2B+2y%5E2+%2B+17y+-6+%3D+0

Answer 6

6x^2 + 3xy + 2y^2 + 17y -6 = 0

\[6*\frac{ d }{ dx }x^2 + 3*[x*\frac{ d }{ dy }y \frac{ dy }{ dx }+y \frac{ d }{ dx }x]+2*\frac{ d }{ dy }y^2\frac{ dy }{ dx }+17*\frac{ d }{ dy }y \frac{ dy }{ dx }-\frac{ d }{ dx }6=\frac{ d }{ dx }0\]

Answer 7

Just have to use the product rule for the x*y term, and the chain rule on each of the Y terms
\[\frac{ d }{ dx } = \frac{ d }{ dy }\frac{ dy }{ dx }\]

Answer 8

given the linked response, how would I find the slope of the tangent line at a given point

Answer 9

after you evaluate , you solve for \[\frac{ dy(x) }{ dx }\]

Answer 10

Then the slope at any value x=a, if the value of y ' (a) or \[\frac{ dy(a) }{ dx } = y~'(a)\]

Answer 11

the derivative you found is the slope of a tangent line at any point of the graph at x= some value a

Answer 12

so, at the point -1,0 , I'd just plug in -1 for the x?

Answer 13

yes, the slope of the tangent line to y(x) at the point (-1,0), will be
\[\frac{ dy(-1) }{ dx } = y~'(-1)\]

Answer 14

The equation for the tangent line to the graph at x=a is,

y - y(a) = y ' (a)*(x - a)

point-slope form for a line, with the slope being y ' (a), the derivative evaluated at that point.

Answer 15

what about the y variables in the derivative? do I plug in 0 for them?

Answer 16

yes, (x,y) = (-1,0)

Answer 17

getting -6/7 for that slope, does that sound right?

Answer 18

oops, it's positive 6/7

Answer 19

I think it will be a positive value.. here is the graph..

Answer 20

thanks for your help

Answer 21

here is it with the tangent line as a check... looks good..