# What is the implicit derivative of #4= (x+y)^2 #?

You can use calculus and spend a few minutes on this problem or you can use algebra and spend a few seconds, but either way you'll get

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To find the implicit derivative of (4 = (x + y)^2), you differentiate both sides of the equation with respect to (x), treating (y) as a function of (x). The derivative of (4) with respect to (x) is (0).

Differentiating ((x + y)^2) with respect to (x) using the chain rule and the power rule, we get:

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

Applying the chain rule to (\frac{d}{dx}(x + y)), we have:

[ \frac{d}{dx}(x + y) = 1 + \frac{dy}{dx} ]

Substituting this back into our expression, we get:

[ 2(x + y)(1 + \frac{dy}{dx}) = 0 ]

Now, solve for (\frac{dy}{dx}):

[ \begin{align*} 2(x + y) \left(1 + \frac{dy}{dx}\right) &= 0 \ 2(x + y) + 2(x + y) \frac{dy}{dx} &= 0 \ 2(x + y) \frac{dy}{dx} &= -2(x + y) \ \frac{dy}{dx} &= \frac{-2(x + y)}{2(x + y)} \ \frac{dy}{dx} &= -1 \end{align*} ]

Therefore, the implicit derivative of (4 = (x + y)^2) with respect to (x) is (-1).

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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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