anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
First step: Take derivative with respect to x on both sides of the equation.
myininaya
  • 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. :) ----
anonymous
  • anonymous
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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

myininaya
  • 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.
myininaya
  • myininaya
\[\text{What did you get when you did } (2xy)' \text{ and } (y^2)' ?\]
anonymous
  • anonymous
(2xy)' = dy/dx 2y +2x dy/dx and (y^2)' = -2y
myininaya
  • myininaya
Looks incorrect for both
myininaya
  • myininaya
\[(2xy)' =2(xy)' \text{ By constant multiple rule}\] \[2(xy)'=2[xy]'=2[(x)'y+x(y)'] \text{ By product rule }\]
myininaya
  • myininaya
\[(y^n)'=n y ^{n-1} y' \text{ By chain rule } \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.