# How do you find the derivative for #y= (x^2 + 2x + 5) / (x + 1)#?

You could differentiate this function by using the quotient rule, or you could simplify the function first, then use the chain rule.

Here's how you could simplify this function. Notice that the numerator can be written as

This is an important thing to notice because you can use the formula for the square of a binomial

to write

You can then go ahead and simplify this by

Your target derivative will thus be equal to

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To find the derivative of y = (x^2 + 2x + 5) / (x + 1), we'll use the quotient rule. The quotient rule states that if you have a function in the form of f(x)/g(x), then the derivative is (g(x) * f'(x) - f(x) * g'(x)) / (g(x))^2. Applying this to the given function:

f(x) = x^2 + 2x + 5 g(x) = x + 1

Now, we'll find the derivatives of f(x) and g(x):

f'(x) = 2x + 2 g'(x) = 1

Now, applying the quotient rule:

y' = [(x + 1)(2x + 2) - (x^2 + 2x + 5)(1)] / (x + 1)^2

Expanding and simplifying:

y' = [2x^2 + 2x + 2x + 2 - x^2 - 2x - 5] / (x + 1)^2 y' = (x^2 + 4x - 3) / (x + 1)^2

So, the derivative of y = (x^2 + 2x + 5) / (x + 1) is y' = (x^2 + 4x - 3) / (x + 1)^2.

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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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