Find an equation of the tangent line to the curve
xey + yex = 5
at the point (0, 5). I tried to derive the slope through implicit differentiation but it says my answer is wrong. Am I not supposed to do that, or is it my math?

Question

Find an equation of the tangent line to the curve 
xey + yex = 5
 at the point (0, 5). I tried to derive the slope through implicit differentiation but it says my answer is wrong. Am I not supposed to do that, or is it my math?

anonymous · Answer

I got $$y \prime=(-ye ^{-x}-e ^{y})/xe ^{y}+e ^{x}$$

anonymous · Answer

then plugged in x and y and got slope of (-5-e^5)

anonymous · Answer

and ended with equation of y=(-5-e^5)x+5

zepdrix · Answer

$$\Large\rm y \prime=(-ye ^{\color{red}{-x}}-e ^{y})/xe ^{y}+e ^{x}$$Why is this negative?

zepdrix · Answer

I think there was a little boo boo there :o

anonymous · Answer

ah i think that was a typo because on my paper i dont have it there

anonymous · Answer

so the answer i got is still the same but it says it is wrong

zepdrix · Answer

oh ok :) ill continue to check then hehe

anonymous · Answer

haha thanks

zepdrix · Answer

Hmm I don't see any mistakes 0_o
It is instructing you to put the line in any specific form? Like point-slope maybe?

anonymous · Answer

oh nvm i figured it out! normally it has the y= on the outside so i don't write it but in this case I had to write the y= part

zepdrix · Answer

o