# 1/x^3+1/y^3=9. (1/2,1)?

differentiate the function ,and write an equation for the tangent line to the graph at the given point

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- mizooLv 74 weeks agoFavorite Answer
-3 * x^(-4) - 3 * y^(-4) * y' = 0

-3 * (1/2)^(-4) - 3 * 1^(-4) * y' = 0

y' = (1/2)^(-4)/1

y' = 1/16

y' = 1/16, (1/2, 1):

y - 1 = (1/16)(x - 1/2)

y = (1/16)x + 31/32

- MyRankLv 64 weeks ago
1/x³ + 1/y³ = 9

x⁻³ + y⁻³ = 9-3x⁻⁴ + (-3) y⁻⁴.dy/dx = 0-3[1/x⁴ + 1/y⁴ .dy/dx] = 01/x⁴ + 1/y⁴ . dy/dx = 0dy/dx = -1/x⁴ x y⁴= -y⁴/x⁴ at (1/2, 1)= -(1)⁴ / (1/2)⁴= -1/1/16dy/dx = -16equation of tangent y - y₁ = m(x-x₁)y-1 = -16(x-1/2).

Source(s): http://myrank.co.in/ - ted sLv 74 weeks ago
[ - 3 / x^4 ] + [ - 3 / y^4] [ dy /dx ] = 0 ===> [ - 48 ] + [ - 3 ] [ dy /dx ] = 0....you finish

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