# Calculus

Use implicit differentiation to find an equation of the tangent line to the graph at the given point.

x + y − 1 = ln(x^5 + y^5), (1, 0)

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1. x + y − 1 = ln(x^5 + y^5)
1 + y' = 1/(x^5+y^5) (5x^4 + 5y^4 y')
y'(1 - 5y^4/(x^5+y^5) = 5x^4/(x^5+y^5) - 1
y' = ((x-5)x^4 - y^5)/(x^5 + (y-5)y^4)

At (1,0),

y' = (-4)/(-1) = 4

So, the line is

y = 4(x-1)

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posted by Steve

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