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

Oct 3, 2017

$0$

#### Explanation:

$\frac{d}{\mathrm{dx}} {\sec}^{- 1} x = \frac{1}{| \quad x | \sqrt{{x}^{2} - 1}}$

$\frac{d}{\mathrm{dx}} \cos e {c}^{- 1} x = - \frac{1}{| \quad x | \sqrt{{x}^{2} - 1}}$

$\therefore$,
$y = {\sec}^{-} 1 \left(2 x\right) + \cos e {c}^{-} 1 \left(2 x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{| \quad 2 x | \sqrt{4 {x}^{2} - 1}} \cdot 2 - \frac{1}{| \quad 2 x | \sqrt{4 {x}^{2} - 1}} \cdot 2$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

ENJOY MATHS !!!!!

Oct 3, 2017

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

#### Explanation:

Given: $y = {\sec}^{-} 1 \left(2 x\right) + {\csc}^{-} 1 \left(2 x\right)$

Differentiate each term:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d \left({\sec}^{-} 1 \left(2 x\right)\right)}{\mathrm{dx}} + \frac{d \left({\csc}^{-} 1 \left(2 x\right)\right)}{\mathrm{dx}} \text{ [1]}$

From any reference table, we know that:

(d(sec^-1(u)))/(du) = 1/(|u|sqrt(u^2-1)

In our case, $u = 2 x$, then $\frac{\mathrm{du}}{\mathrm{dx}} = 2$

Using the chain rule:

$\frac{d \left({\sec}^{-} 1 \left(2 x\right)\right)}{\mathrm{dx}} = \frac{2}{| 2 x | \sqrt{4 {x}^{2} - 1}} \text{ [2]}$

Similarly, we know that:

(d(csc^-1(u)))/(du) = -1/(|u|sqrt(u^2-1)

In our case, $u = 2 x$, then $\frac{\mathrm{du}}{\mathrm{dx}} = 2$

Using the chain rule:

$\frac{d \left({\csc}^{-} 1 \left(2 x\right)\right)}{\mathrm{dx}} = - \frac{2}{| 2 x | \sqrt{4 {x}^{2} - 1}} \text{ [3]}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{| 2 x | \sqrt{4 {x}^{2} - 1}} - \frac{2}{| 2 x | \sqrt{4 {x}^{2} - 1}}$

The terms sum to 0:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$