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

Explanation:

Step1: Recall derivative formulas

Recall that $\frac{d}{dx}(\tan^{-1}u)=\frac{1}{1 + u^{2}}\cdot\frac{du}{dx}$ and $\frac{d}{dx}(\csc^{-1}x)=-\frac{1}{|x|\sqrt{x^{2}-1}}$. Let $u = \sqrt{x^{2}-1}$.

Step2: Differentiate $\tan^{-1}\sqrt{x^{2}-1}$

First, find $\frac{du}{dx}=\frac{2x}{2\sqrt{x^{2}-1}}=\frac{x}{\sqrt{x^{2}-1}}$. Then $\frac{d}{dx}(\tan^{-1}\sqrt{x^{2}-1})=\frac{1}{1+(x^{2}-1)}\cdot\frac{x}{\sqrt{x^{2}-1}}=\frac{x}{x^{2}\sqrt{x^{2}-1}}=\frac{1}{x\sqrt{x^{2}-1}}$.

Step3: Differentiate $-\csc^{-1}x$

Using the formula $\frac{d}{dx}(\csc^{-1}x)=-\frac{1}{|x|\sqrt{x^{2}-1}}$, the derivative of $-\csc^{-1}x$ is $\frac{1}{|x|\sqrt{x^{2}-1}}$.

Step4: Combine the derivatives

Since $y = \tan^{-1}\sqrt{x^{2}-1}-\csc^{-1}x$, then $\frac{dy}{dx}=\frac{1}{x\sqrt{x^{2}-1}}+\frac{1}{|x|\sqrt{x^{2}-1}}$. Given $x > 1$, $|x|=x$, and $\frac{1}{x\sqrt{x^{2}-1}}+\frac{1}{x\sqrt{x^{2}-1}}=\frac{2}{2x^{2}\sqrt{x^{2}-1}}+\frac{1}{x\sqrt{x^{2}-1}}$.

Answer:

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