# How do I find the derivative of #y = arc cot(x) + (cot x)^-1#?

First, start by rewriting your function as

This is equivalent to

Use the trigonometric identity

to write

which means that the derivative will be

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To find the derivative of ( y = \text{arccot}(x) + (\cot x)^{-1} ), you differentiate each term separately using the chain rule and the power rule for derivatives:

[ \frac{d}{dx} \left( \text{arccot}(x) + (\cot x)^{-1} \right) = \frac{d}{dx} \left( \text{arccot}(x) \right) + \frac{d}{dx} \left( (\cot x)^{-1} \right) ]

[ = -\frac{1}{1 + x^2} + \frac{d}{dx} \left( \frac{1}{\cot x} \right) ]

[ = -\frac{1}{1 + x^2} - \frac{d}{dx} \left( \tan x \right) ]

[ = -\frac{1}{1 + x^2} - \frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) ]

[ = -\frac{1}{1 + x^2} - \frac{\cos^2 x + \sin^2 x}{\cos^2 x} ]

[ = -\frac{1}{1 + x^2} - \frac{1}{\cos^2 x} ]

[ = -\frac{1}{1 + x^2} - \sec^2 x ]

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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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