Question

How would you use implicit differentiation to find an equation of the tangent line to the curve

x^2+ 2xy-y^2+x=2 at the point (1,2)

Answer 1

First step: Take derivative with respect to x on both sides of the equation.

Answer 2

\[ (x^2+2xy-y^2+x)'=(2)'\]
-----
\[(x^2+2xy-y^2+x)'=(x^2)'+(2xy)'-(y^2)'+(x)'=....\]
\[\text{Here you will need power rule for } (x^2)'\]
\[\text{You will need constant multiple rule along with product rule for } (2xy)'\]
\[\text{You will need chain rule for } (y^2)'\]
\[\text{You will need to know that }(x)'=1 \text{ for the } (x)' \text{part}\]

----
\[(2)'=....\]

You should know the derivative of a constant. :)
----

Answer 3

i understand the steps i just cannot put 2 and 2 together and get the correct final answer. i got 1/3 as the slope, is the correct?

Answer 4

Hmm...Not getting 1/3....

Show me your work so I can see where you messed up.
You can do it one line at a time if you want.

Answer 5

\[\text{What did you get when you did } (2xy)' \text{ and } (y^2)' ?\]

Answer 6

(2xy)' = dy/dx 2y +2x dy/dx and (y^2)' = -2y

Answer 7

Looks incorrect for both

Answer 8

\[(2xy)' =2(xy)' \text{ By constant multiple rule}\]
\[2(xy)'=2[xy]'=2[(x)'y+x(y)'] \text{ By product rule }\]

Answer 9

\[(y^n)'=n y ^{n-1} y' \text{ By chain rule } \]