# How do you find #(dy)/(dx)# given #(2y-x)^2+3x=0#?

Differentiate each term:

The first term requires the use of the chain rule:

Substituting into equation [2]:

Reverse the substitution for g:

Substitute back into equation [1]:

The derivative of the second term is trivial and the derivative of 0 is 0:

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To find ( \frac{{dy}}{{dx}} ) given ( (2y - x)^2 + 3x = 0 ), differentiate both sides of the equation implicitly with respect to ( x ), then solve for ( \frac{{dy}}{{dx}} ). Differentiate the equation term by term, using chain rule where necessary.

Starting with ( (2y - x)^2 + 3x = 0 ), differentiate each term:

- Differentiate ( (2y - x)^2 ) with respect to ( x ) using chain rule.
- Differentiate ( 3x ) with respect to ( x ).

After differentiation, solve for ( \frac{{dy}}{{dx}} ).

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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.

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.

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.

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