Asked by Matt
Find the equation of the tangent line to the curve 2e^(xy) = x + y at (0,2)
Answers
Answered by
Reiny
differentiate implicity with respect to x
2e^(xy)*(xdy/dx + y) = 1 + dy/dx
expanding the left side,
2e^(xy)xdy/dx + 2e^(xy)y - dy/dx = 1
taking out a common factor of dy/dx and solving for that gives us
dy/dx = (1 - 2e^(xy)y)/(2e^(xy)x - 1)
at (0,2)
dy/dx = (1 - 2e^0(2))/(2e^0(0) - 1)
= 3
since (0,2) is the y-intercept and the slope is 3 we can just write the equation of the tangent as
y = 3x + 2
2e^(xy)*(xdy/dx + y) = 1 + dy/dx
expanding the left side,
2e^(xy)xdy/dx + 2e^(xy)y - dy/dx = 1
taking out a common factor of dy/dx and solving for that gives us
dy/dx = (1 - 2e^(xy)y)/(2e^(xy)x - 1)
at (0,2)
dy/dx = (1 - 2e^0(2))/(2e^0(0) - 1)
= 3
since (0,2) is the y-intercept and the slope is 3 we can just write the equation of the tangent as
y = 3x + 2
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