How do you implicitly differentiate #x^3 y-3xy^2# at the point (1,-10)?

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Answer 1 · 2015-08-14T09:46:41

Refer Explanation

#x^3#y - 3#y^2# = 0

To differentiate #x^3#y use product rule

#x^3#.#dy/dx# + y.3#x^2# - 6y#dy/dx# = 0

Rewrite it -

#x^3#.#dy/dx# + 3#x^2#y - 6y#dy/dx# = 0

Keep #dy/dx# terms on the left hand side

#x^3#.#dy/dx# - 6y#dy/dx# = - 3#x^2#y

Take #dy/dx# as common factor

#dy/dx#(#x^3# - 6y) = - 3#x^2#y
Solve it for #dy/dx#

#dy/dx# = #( - 3x^2y)/(x^3- 6y)#

At x = 1 and Y = - 10

#dy/dx# = #( - 3(1^2)( - 10))/(1^3- 6( - 10))# = #30/61#