# What is the equation of the line tangent to # f(x)=1/ln(x^2+x)*(x^2+x) # at # x=-1 #?

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To find the equation of the line tangent to the function f(x) at x = -1, we need to find the derivative of f(x) and evaluate it at x = -1.

First, let's find the derivative of f(x). Using the product rule and the chain rule, we have:

f'(x) = [(1/ln(x^2+x))*(x^2+x)]' = [(1/ln(x^2+x))]'*(x^2+x) + (1/ln(x^2+x))*[(x^2+x)']

To simplify further, we need to find the derivatives of (1/ln(x^2+x)) and (x^2+x).

The derivative of (1/ln(x^2+x)) can be found using the chain rule:

[(1/ln(x^2+x))]' = -[(1/ln(x^2+x))^2] * (ln(x^2+x))' = -[(1/ln(x^2+x))^2] * (1/(x^2+x)) * (x^2+x)'

The derivative of (x^2+x) is simply 2x + 1.

Now, let's substitute these derivatives back into the equation for f'(x):

f'(x) = -[(1/ln(x^2+x))^2] * (1/(x^2+x)) * (2x + 1) * (x^2+x) + (1/ln(x^2+x))*[(2x + 1)]

Next, we can evaluate f'(x) at x = -1:

f'(-1) = -[(1/ln((-1)^2+(-1)))^2] * (1/((-1)^2+(-1))) * (2(-1) + 1) * ((-1)^2+(-1)) + (1/ln((-1)^2+(-1)))*[(2(-1) + 1)]

Simplifying this expression will give us the slope of the tangent line at x = -1.

Finally, we can use the point-slope form of a line to find the equation of the tangent line. The equation will have the form y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope we found earlier.

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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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