Question

ind the eqn of tangent line to
x^2 + 2xy - y^2 + x = 2 at (1,2)

Im not sure how you would isolate y and take the derivative

Answer 1

this is a quadreatic in y solve using the quadratic equation. i'll try to get an answer to you. computer is very slow

Answer 2

first solve the initial quadratic in y , (-y^2 +2xy-(2-x+x^2)=0)
which should give you
-x +- root(2x^2-x+2) =y

you should be able to differentiate it from here knowing that a root is the same as to the power of a half.

Answer 3

\[2x+2y+2x \frac{dy}{dx}-2y \frac{dy}{dx}+1=0\]
\[2x+2y+1=\frac{dy}{dx} (-2x+2y)\]
\[\frac{dy}{dx}=\frac{2x+2y+1}{-2x+2y}\]

Answer 4

Now plug in (1,2)
\[\frac{2(1)+2(2)+1}{-2(1)+2(2)}=\frac{7}{2}\]

Answer 5

If I remember implicit differentiation correctly haha.

Answer 6

@jf55 You don't have to isolate the y to differentiate it. You can do whats called implicit. Think about this, if you have y=f(x)
And you differentiate it with respect to x, you get:
\[\frac{dy}{dx}=f'(x)\]
Its the same thing if you have a y in the function.
\[\frac{d}{dx}(x^2y+2x+y^3)=2xy+x^2 \frac{dy}{dx}+2+3y^2 \frac{dy}{dx}\]
(NOTE: the first one you have to use a product rule)

Answer 7

@jf55 You don't have to isolate the y to differentiate it. You can do whats called implicit. Think about this, if you have y=f(x)
And you differentiate it with respect to x, you get:
\[\frac{dy}{dx}=f'(x)\]
Its the same thing if you have a y in the function.
\[\frac{d}{dx}(x^2y+2x+y^3)=2xy+x^2 \frac{dy}{dx}+2+3y^2 \frac{dy}{dx}\]
(NOTE: the first one you have to use a product rule)

Answer 8

http://tutorial.math.lamar.edu/Classes/CalcI/ImplicitDIff.aspx