# How do you differentiate #y=sec^-1(2x)+csc^-1(2x)#?

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Differentiate each term, using table lookup and the chain rule.

Combine the terms.

Differentiate each term:

From any reference table, we know that:

Using the chain rule:

Similarly, we know that:

Using the chain rule:

Substitute equations [2] and [3] into equation [1]:

The terms sum to 0:

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To differentiate ( y = \sec^{-1}(2x) + \csc^{-1}(2x) ):

- Use the chain rule for differentiation.
- Differentiate each term separately.
- For ( \sec^{-1}(2x) ), differentiate ( \sec^{-1}(u) ) with respect to ( u ) and then multiply by the derivative of ( u = 2x ).
- For ( \csc^{-1}(2x) ), differentiate ( \csc^{-1}(v) ) with respect to ( v ) and then multiply by the derivative of ( v = 2x ).

The derivatives of the inverse trigonometric functions are as follows:

- ( \frac{d}{du}(\sec^{-1}u) = \frac{1}{|u|\sqrt{u^2-1}} )
- ( \frac{d}{dv}(\csc^{-1}v) = -\frac{1}{|v|\sqrt{v^2-1}} )

Apply the chain rule and simplify the derivatives accordingly.

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