Question

Use implicit differentiation to find the slope of the tangent line to the curve

y/x+6y=x^2–6 at the point (1,–5/31) .

Again i think i'm messing up with the algebra here. I used quotient rule to get
[(x+6y)(y')-(y)(1+6y')]/(x+6y)^2=2x

I don't know how to go from here to find the m. Please help :)

Answer 1

Your derivative equation looks good.

sub in x=1 and y = -5/31 into that derivative equation, then solve it for y'
that will be your m
Good luck with that messy arithmetic.

Answer 2

I got y'=(2x+x^2+12yx+6y^2-y)/(x+6y+6)
But when i plugged in the numbers it came out right... do you think the equation is wrong or i just typed in the numbers wrong?

BTW thanks for replying!

Answer 3

using your [(x+6y)(y')-(y)(1+6y')]/(x+6y)^2=2x
we can continue
xy' + 6yy' - y - 6yy' = 2x
xy' - y = 2x
y' = (2x+y)/x
using the given point:
y' = (2 - 5/31)/1 = 57/31

y + 5/31 = (57/31)(x-1)
31y + 5 = 57x - 57
57x - 31y = 62

check my arithmetic.

Answer 4

I did this exactly with this question and i got the wrong answer.

Use implicit differentiation to find the slope of the tangent line to the curve

y/(x+5y)=x^3+5
at the point (x=1, y=6/–29).

But I noticed after "we can continue..."
the denominator was moved to the right side of the equation.. so maybe that's it..??

Answer 5

Sorry. I meant the denominator from the quotient rule was never multiplied to the right or so. I don't see it. Where did that portion go?