A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Find an equation for the tangent line to the curve x^3+xy+y^3=x at the point (1,0)
Would the first step be x^3+y+x+3y^2=x
anonymous
 5 years ago
Find an equation for the tangent line to the curve x^3+xy+y^3=x at the point (1,0) Would the first step be x^3+y+x+3y^2=x

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, you need to take the derivative, you have to take the derivative of each item, not just the y's: \[d/dx(x^3+xy+y^3)=d/dx(x)\] \[3x^2 + 1y + xy' + 3y^2y'=1\] \[xy'+3y^2y'=13x^2y\] \[y'(x+3y^2)=13x^2y\] \[y'=(13x^2y)/(x+3y^2)\] Now, put (1, 0) into y' to get the slope of the tangent line. Put a 1 where you see an x and a 0 where you see a y. y' = (1  3(1)^20)/(1+3(0)^2) = 2/1 = 2 Now that you have a slope m = 2, use that with the point (1, 0) and get an equation in pointslope form: y  0 = 2(x  1) > y = 2x + 2
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.