Find equatin of the tangent line(implicit diff)

Question

Hi, I just need to know if my answer is correct. Its asking to find the equation of the tangent line for : x²e^y +ye^x =4 at (2,0) 

 

 

Answer: y=-x/2 +1

Answer 1

Using implicit differentiation, I get

 

 dy/dx = -(2 x e^y + y e^x) /(x² e^y + e^x)

 

 Evaluating this for x = 2, y = 0  gives

 

   m =  dy/dx =  -4/(4 + e²)

 

  Using the point slope form, the equation of the tangent line becomes

 

   y - 0 =  -4/(4+ e²)  ( x -2)

 

 

 

 

Answer 2

The slope of the tangent line has to be y'.  Let us start by evaluating y'.

 

x²e^y + ye^x = 4

 

Take d/dx () on both sides to get y'.  Use the product rule with the power, exponential, and chain rules.

 

2xe^y + x²e^yy' + y'e^x + ye^x = 0

 

y'(x²e^y + e^x) = - (2xe^y + ye^x)

 

y' = - (2xe^y + ye^x) / (x²e^y + e^x)

 

At (2,0) this gives

 

y' = - (2*2e⁰ + 0*e²) / (2²e⁰ + e²)

 

y' = - 4 / (4 + e²)

 

You got a slope of -1/2.  Double check the math (both yours and mine!).  But once you get a slope m, the equation is

 

y = mx + b

 

and get b from (2,0).

 

0 = m(2) + b