## youngbuck2123 3 years ago 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)

1. myininaya

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

2. myininaya

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

3. youngbuck2123

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?

4. myininaya

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.

5. myininaya

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

6. youngbuck2123

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

7. myininaya

Looks incorrect for both

8. myininaya

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

9. myininaya

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