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

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 :)

- Implicit Differentiation -
**Reiny**, Saturday, February 18, 2012 at 10:09pm
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.

- Implicit Differentiation -
**jasmineT**, Saturday, February 18, 2012 at 10:18pm
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!

- Implicit Differentiation -
**Reiny**, Saturday, February 18, 2012 at 10:32pm
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.

- Implicit Differentiation -
**Lisa**, Sunday, May 19, 2013 at 4:07am
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..??

- Implicit Differentiation -
**Lisa**, Sunday, May 19, 2013 at 4:09am
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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