How do you find an equation of the tangent line to the curve #xe^y+ye^x = 1# at the point (0,1)?

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Answer 1 · 2018-03-05T17:42:10

#y=-[e+1]x+1#

Given, #xe^y+ye^x=1# We need to differentiate both sides implicitly with respect to x using the product and chain rule.

Product rule #d/dx[uv]=vdu/dx+udv/dx# where v and u are both
functins of #x#.

So differentiating implicitly both sides, # d/dx[ xe^y+ye^x]= e^y+xe^ydy/dx+e^xdy/dx+ye^x#=[ differential of a constant is zero]

Factoring, collecting like terms and tidying up.....

#dy/dx[xe^y+e^x]=-[e^y+ye^x]# and so #dy/dx=-[e^y+ye^x]/ [xe^y+e^x]# and substituting in the values for x and y..ie.[0,1

#dy/dx=-[e^1+[1]e^0]/[[0]e^1+e^0]#=#-[e+1]/1#=#-[e+1] ie. # [the gradient]#.

Equation of tangent line is #[y-y1]=[m[x-x1]]# where 'm' is the gradient, and so we have [ from the given coordinates]

#y-1=-[e+1][x-0]# and so #y=-[e+1]x+1#.