Question

Use implicit differentiation to find an equation of a tangent line to the curve at a given point?

x^2 + 2xy - y^2 + x = 2
(1,2) (hyperbola)

I got:
2x + 2 (dy/dx) - 2y (dy/dx) = 0
(2 - 2y) dy/dx = -2x
dy/dx = -x / (1-y)

At (1,2) the slope is:
-(1)/(1-(2))=-1/-1=1

y-y1=m(x-x1)
(y-2)=1(x-1)
(y-2)=1x-1
y-2=1x-1
y = x + 1

can someone verify the correct answer?

Answer 1

when using implicit differentiation on 2xy, you have to apply the product rule. \[\frac{d}{dx}2xy=2y+2x*\frac{dy}{dx}\]

Answer 2

Actually the answer is 2y-7x+3=0

Answer 3

yep he mentioned it you made your mistake at that point.

Answer 4

can someone show me the steps then.. so I can see what I did wrong?

Answer 5

you also forgot to apply chain rule on -y^2. \[\frac{d}{dx}-y^2=-2*y*\frac{dy}{dx}\]
the whole derivative should look like this \[2x+2y+2xy'-2y*y'+1=0\]
I also noticed you didn't differentiate the last x.

Try to go back and differntiate everything again and get the rigth answer before moving on to understand it.

Answer 6

2x+2(xy'(x)+y)-2yy'(x)+1=0
2x+2xy'(x)'+2y-2yy'(x)+1=0
2xy'(x)-2yy'(x)=2x-2y-1
y'(x)(2x-2y)=2x-1
y'(x)=(2x-2y-1)/(2x-2y)

(2(1)-2(2)-1)/(2(1)-2(2))
7/2

y=7/2x-3/2

Answer 7

That is correct

Answer 8

yay.. gosh dang lol.