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How do you find #(dy)/(dx)# given #-x^3y^2+4=5x^2+3y^3#?

Answer 1

#dy/dx=-(10x +3x^2y^2)/(2yx^3+ 9y^2) #

#-x^3y^2 + 4= 5x^2+3y^3# #" "# #rArr-3x^2y^2+2ydy/dxxx(-x^3)=10x + 9y^2xxdy/dx# #" "# #rArr-3x^2y^2-2yx^3dy/dx=10x + 9y^2xxdy/dx# #" "# #rArr-2yx^3dy/dx- 9y^2xxdy/dx=10x +3x^2y^2 # #" "# #rArrdy/dx(-2yx^3- 9y^2)=10x +3x^2y^2 # #" "# #rArrdy/dx=(10x +3x^2y^2)/(-2yx^3- 9y^2) # #" "# #rArrdy/dx=-(10x +3x^2y^2)/(2yx^3+ 9y^2) #

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Answer 2

Isabella Ackerman

To find ( \frac{dy}{dx} ) given the equation ( -x^3y^2 + 4 = 5x^2 + 3y^3 ), follow these steps:

Implicitly differentiate the given equation with respect to ( x ).
Apply the chain rule and product rule as needed when differentiating terms involving ( y ).
Solve the resulting equation for ( \frac{dy}{dx} ) to express it explicitly in terms of ( x ) and ( y ).

This process will yield the derivative ( \frac{dy}{dx} ) in terms of ( x ) and ( y ).

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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.