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.
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 :)
5 answers
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!
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.
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..??
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?
1 answer