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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