Posted by jasmineT on Saturday, February 18, 2012 at 10:05pm.
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.
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!
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.
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..??
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?
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