How do you find #(df)/dy# and #(df)/dx# of #f(x,y)=(3x^2-2e^y)/(2x+y)#, using the quotient rule?

Answer 1

Quotient rule states that #d/dx(h/g) = (gh' - hg')/(g^2) #

That can be used independently for each of the variables x and y:

#f(x, y) = (h(x, y)) / g(x, y) # where #h(x, y) = 3x^2 - 2e^y # and #g(x, y) = 2x+y#

Therefore, #(df)/dx = (g cdot (dh)/dx - h cdot (dg)/dx )/(g^2) # #= ((2x+y)(6x) - (3x^2 - 2e^y) (2))/(2x+y)^2 = (6x^2+6xy + 4e^y)/(2x+y)^2#

#(df)/dy = (g cdot (dh)/dy - h cdot (dg)/dy )/(g^2) # #= ((2x+y)(-2e^y) - (3x^2 - 2e^y) (1))/(2x+y)^2 = (-4xe^y - 2ye^y - 3x^2 + 2e^y)/(2x+y)^2#

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Answer 2

Emily Ammerman

To find (df)/dy and (df)/dx of f(x,y)=(3x^2-2e^y)/(2x+y) using the quotient rule:

(df)/dx = [(2x+y)(d/dx)(3x^2-2e^y) - (3x^2-2e^y)(d/dx)(2x+y)] / (2x+y)^2

(df)/dy = [(2x+y)(d/dy)(3x^2-2e^y) - (3x^2-2e^y)(d/dy)(2x+y)] / (2x+y)^2

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.