Question

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.31. y = cos⁻¹(x) - x sech⁻¹(x)

Accepted Answer

Identify the function to differentiate: $$y = \cos$$^{-1}$$(x) - x \ 	ext{sech}^{-1}(x)$$, where $$\cos$$^{-1}$$(x) is $$the inverse cosine function and $$	ext{sech}^{-1}(x) is $$the inverse hyperbolic secant function. Recall the derivative formulas for the inverse functions involved: 
- For $$y = \cos$$^{-1}$$(x)$$, the derivative is $$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2$}}.
- $$For $$y = 	ext{sech}^{-1}(x)$$, the derivative is $$\frac{dy}{dx} = -\frac{1}{x \sqrt{1 - x^2$}} ($$valid for $$0 \u003c x \u003c 1). $$Apply the product rule to the term $$x \ 	ext{sech}^{-1}(x)$$: 
If $$u = x$$ and $$v = 	ext{sech}^{-1}(x)$$, then $$\frac{d}{dx}(uv) = u'v + uv'.
$$Calculate $$u' = 1$$ and use the derivative of $$v$$ from the previous step. Write the derivative of $$y as$$:
$$\frac{dy}{dx} = \frac{d}{dx} \left( \cos$$^{-1}$$(x) ight) - \frac{d}{dx} \left( x \ 	ext{sech}^{-1}(x) ight)$$,
which becomes
$$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2$}} - \left( 1 \cdot 	ext{sech}^{-1}(x) + x \cdot \left(-\frac{1}{x \sqrt{1 - x^2$}}ight) ight). $$Simplify the expression by distributing the negative sign and combining like terms, keeping the derivative in terms of $$x$$ without evaluating the final numeric value.

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
31. y = cos⁻¹(x) - x sech⁻¹(x)

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Key Concepts

Derivative of Inverse Trigonometric Functions

Derivative of Inverse Hyperbolic Functions

Product Rule for Differentiation