find the derivative of y with respect to x.\ny = tan⁻¹√(x² - 1) - csc⁻¹x, x > 1\ndy/dx = 2/(x√(x² - 1))

Explanation:

Step1: Differentiate $\tan^{-1}\sqrt{x^{2}-1}$

Let $u = \sqrt{x^{2}-1}$, then $\frac{d}{dx}(\tan^{-1}u)=\frac{1}{1 + u^{2}}\cdot\frac{du}{dx}$. First, $\frac{du}{dx}=\frac{2x}{2\sqrt{x^{2}-1}}=\frac{x}{\sqrt{x^{2}-1}}$, and $1 + u^{2}=1+(x^{2}-1)=x^{2}$. So $\frac{d}{dx}(\tan^{-1}\sqrt{x^{2}-1})=\frac{1}{x^{2}}\cdot\frac{x}{\sqrt{x^{2}-1}}=\frac{1}{x\sqrt{x^{2}-1}}$.

Step2: Differentiate $-\csc^{-1}x$

The derivative of $\csc^{-1}x$ is $-\frac{1}{|x|\sqrt{x^{2}-1}}$, since $x > 1$, then $\frac{d}{dx}(-\csc^{-1}x)=\frac{1}{x\sqrt{x^{2}-1}}$.

Step3: Use sum - rule of derivatives

$\frac{dy}{dx}=\frac{d}{dx}(\tan^{-1}\sqrt{x^{2}-1})+\frac{d}{dx}(-\csc^{-1}x)$. Substitute the results from step 1 and step 2: $\frac{dy}{dx}=\frac{1}{x\sqrt{x^{2}-1}}+\frac{1}{x\sqrt{x^{2}-1}}=\frac{2}{x\sqrt{x^{2}-1}}$.

Answer:

$\frac{2}{x\sqrt{x^{2}-1}}$