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What is the implicit derivative of #25=yx^2-xy+y^2x#?

Answer 1

Aurora Alvarado

Using the chain and product rules ...

#25=yx^2-xy+y^2x#

Now using implicit differentiation ...

#0=y'(x^2)+y(2x)-y-xy'+2yy'x+y^2#

Next, solve for #y'#

#y'(x^2-x+2xy)=-(2xy-y+y^2)#

#y'=-(2xy-y+y^2)/(x^2-x+2xy)#

hope that helped

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Answer 2

Charles Anctil

To find the implicit derivative of (25 = yx^2 - xy + y^2x), differentiate both sides of the equation with respect to (x), treating (y) as a function of (x).

The implicit derivative is:

[\frac{dy}{dx} = \frac{d}{dx}(yx^2) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^2x)]

Using the product rule and chain rule, we get:

[\frac{dy}{dx} = x^2\frac{dy}{dx} + 2xy - y - x\frac{dy}{dx} + 2yx\frac{dy}{dx} + y^2]

Rearranging terms, we get:

[\frac{dy}{dx}(x^2 - x + 2xy) = y - 2xy - y^2]

So, the implicit derivative is:

[\frac{dy}{dx} = \frac{y - 2xy - y^2}{x^2 - x + 2xy}]

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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.