How do you implicitly differentiate #3=5x^3yx^2y+y^2/x#?
y'=
d/dx#( 5x^3y)d/dx(x^2y)+d/dx(y^2/x)#=d/dx(3)
Use product rule for first two and quotient rule for third part
A rational expression is 0 , only if the numerator is 0
solve for y'
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To implicitly differentiate the equation (3 = 5x^3y  x^2y + \frac{y^2}{x}) with respect to (x), follow these steps:
 Differentiate each term of the equation with respect to (x).
 Apply the product rule and quotient rule where necessary.
 Solve for (\frac{dy}{dx}).
Here are the steps in detail:
Given: ( 3 = 5x^3y  x^2y + \frac{y^2}{x} )

Differentiate each term with respect to (x): [ \frac{d}{dx}(3) = \frac{d}{dx}(5x^3y)  \frac{d}{dx}(x^2y) + \frac{d}{dx}\left(\frac{y^2}{x}\right) ] (0 = 15x^2y + 5x^3\frac{dy}{dx}  2xy  x^2\frac{dy}{dx}  \frac{y^2}{x^2})

Combine like terms: (0 = (15x^2y  2xy) + (5x^3  x^2)\frac{dy}{dx}  \frac{y^2}{x^2})

Solve for (\frac{dy}{dx}): (\frac{dy}{dx} = \frac{\frac{y^2}{x^2}  (15x^2y  2xy)}{5x^3  x^2})
Thus, the implicit derivative of the given equation with respect to (x) is: [ \frac{dy}{dx} = \frac{\frac{y^2}{x^2}  15x^2y + 2xy}{5x^3  x^2} ]
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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