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How do you differentiate #y=x^2y-y^2x#?

Answer 1

I found: #(dy)/(dx)=(2xy-y^2)/(-x^2+2yx+1)#

We use Implicit Differentiation where we remember that #y# is a function of #x# and must be derived accordingly; for example if you have #y^2# you derive it getting: #2y(dy)/(dx)# to take into account this depndence.

In your case we get:

#1(dy)/(dx)=2xy+x^2(dy)/(dx)-2yx(dy)/(dx)-y^2# collect #(dy)/(dx)# and rearrange: #(dy)/(dx)=(2xy-y^2)/(-x^2+2yx+1)#

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Answer 2

Charles Arocha

To differentiate the given expression, ( y = x^2y - y^2x ), with respect to ( x ), you can use the product rule and the chain rule:

Apply the product rule to ( x^2y ) and ( -y^2x ). [ \frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx} ] [ \frac{d}{dx}(-y^2x) = -y^2 - 2xy\frac{dy}{dx} ]
Combine the results. [ \frac{d}{dx}(y) = 2xy + x^2\frac{dy}{dx} - y^2 - 2xy\frac{dy}{dx} ]
Simplify the expression. [ \frac{d}{dx}(y) = (2xy - 2xy)\frac{dy}{dx} + (x^2 - y^2) ]
Further simplify to get the final answer. [ \frac{d}{dx}(y) = (x^2 - 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.