# How do you find the derivative of # y = cot5x + csc5x #?

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To find the derivative of ( y = \cot^5(x) + \csc^5(x) ), you can use the chain rule and the derivatives of cotangent and cosecant functions:

[ \frac{dy}{dx} = \frac{d}{dx}(\cot^5(x)) + \frac{d}{dx}(\csc^5(x)) ]

Applying the chain rule and the power rule for derivatives:

[ \frac{dy}{dx} = 5\cot^4(x)\frac{d}{dx}(\cot(x)) - 5\csc^4(x)\frac{d}{dx}(\csc(x)) ]

Using the derivatives of cotangent and cosecant:

[ \frac{dy}{dx} = 5\cot^4(x)(-\csc^2(x)) + 5\csc^4(x)(-\cot(x)\csc(x)) ]

[ \frac{dy}{dx} = -5\cot^4(x)\csc^2(x) + (-5)\csc^5(x)\cot(x) ]

[ \frac{dy}{dx} = -5\cot^4(x)\csc^2(x) - 5\csc^5(x)\cot(x) ]

So, the derivative of ( y = \cot^5(x) + \csc^5(x) ) is ( \frac{dy}{dx} = -5\cot^4(x)\csc^2(x) - 5\csc^5(x)\cot(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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