How do you implicitly differentiate #y= xy^2 + x ^2 e^(4x+y) #?
See answer below:
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To implicitly differentiate ( y = xy^2 + x^2 e^{4x+y} ):
 Differentiate each term with respect to ( x ).
 Apply the product rule for the term ( xy^2 ).
 Apply the product rule and chain rule for the term ( x^2 e^{4x+y} ).
 Solve for ( \frac{dy}{dx} ) after differentiation.
The steps are as follows:

Differentiate each term:
( \frac{dy}{dx} = y^2 + 2xy \frac{dy}{dx} + 2xe^{4x+y} + x^2 \left(4e^{4x+y} + e^{4x+y} \frac{dy}{dx}\right) )

Rearrange terms involving ( \frac{dy}{dx} ):
( \frac{dy}{dx}  2xy \frac{dy}{dx}  x^2 e^{4x+y} \frac{dy}{dx} = y^2 + 2xe^{4x+y} + 4x^2 e^{4x+y} )

Factor out ( \frac{dy}{dx} ):
( \frac{dy}{dx}(1  2xy  x^2 e^{4x+y}) = y^2 + 2xe^{4x+y} + 4x^2 e^{4x+y} )

Solve for ( \frac{dy}{dx} ):
( \frac{dy}{dx} = \frac{y^2 + 2xe^{4x+y} + 4x^2 e^{4x+y}}{1  2xy  x^2 e^{4x+y}} )
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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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