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 $y=-[e+1]x+1$ .

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