How do you differentiate #5x^3-4xy-2y^2=1#?

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Answer 1 · 2015-02-19T17:59:20

Hello,

You can differentiate a function, not an equality.

If you want differentiate a 2-variables function #(x,y)\mapsto f(x,y)#, you write

#d f = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy#.

If #f(x,y)= 5x^3-4xy-2y^2#, you have

# \frac{\partial f}{\partial x} = 15x^2-4y# and # \frac{\partial f}{\partial y} = -4x - 4y#

therefore, you finally have

#d f = (15x^2-4y)d x - 4(x+y)d y#