February 6th, 2012, 06:06 AM  #1 
Member Joined: Jan 2012 Posts: 82 Thanks: 0  Equation of the tangent line
Hi! I need to find dy/dx explicitly (i.e. in the form dy/dx=g(x,y) if x^3+y^2+2y=2. Then I need to determine the equation of the tangent to the curve represented by the above equation at the point (1,1). Can anyone help? Thanks 
February 6th, 2012, 06:30 AM  #2 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,767 Thanks: 5  Re: Equation of the tangent line
This is called "implicit differentiation", since you haven't/won't solve for y explicitly before differentiating. x^3+y^2+2y=2 We are going to use the chain rule on EVERY TERM. I will write dx/dx, just so you can see that the chain rule is always in play. However, dx/dx is just equal to 1, so we will make the appropriate simplification... (x^3) ' + (y^2) ' + (2y) ' = (2) ' ......here I have just put a ' after every term to indicate that we are differentiating. (3x^2 dx/dx) + (2y dy/dx) + (2 dy/dx) = 0 ... chain rule, power rule, and/or constant rule Factor out the dy/dx and simplify... (3x^2) + (dy/dx)*(2y + 2) = 0 Solve for dy/dx... (dy/dx)*(2y + 2) = 3x^2 after dividing. Plug in x = 1, y = 1 to THIS formula to get the slope at the point (1, 1). Then use point slope formula to get an equation of the tangent line. 
February 6th, 2012, 06:36 AM  #3 
Member Joined: Jan 2012 Posts: 82 Thanks: 0  Re: Equation of the tangent line
Thanks, I thought that. Can you please help me with the point slope formula. 
February 6th, 2012, 07:05 AM  #4 
Member Joined: Jan 2012 Posts: 82 Thanks: 0  Re: Equation of the tangent line
Actually Is the following correct, plug in x=1 and y=1, i get the slope m = 3/4 using the point slope fomula i get yy1=m(xx1) y1=3/4(x1) y=3/4x+1/4 
February 6th, 2012, 07:24 AM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Equation of the tangent line
Yes, that is correct. 
February 9th, 2012, 01:29 AM  #6 
Member Joined: Jan 2012 Posts: 82 Thanks: 0  Re: Equation of the tangent line
Thank you for your help 

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equation, line, tangent 
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