Question

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.29. y = (1 - t)coth⁻¹(√t)

Accepted Answer

Identify the function given: $$y = (1 - t) \, 	ext{coth}^{-1}(\sqrt{t}). We $$need to find $$\frac{dy}{dt}$$, the derivative of $$y$$ with respect to $$t. $$Recognize that $$y is a $$product of two functions of $$t$$: $$u = (1 - t)$$ and $$v = 	ext{coth}^{-1}(\sqrt{t}). $$Use the product rule for differentiation: $$\frac{dy}{dt} = u'v + uv'. $$Compute $$u' = \frac{d}{dt}(1 - t) = -1. $$Find $$v' = \frac{d}{dt} \left( 	ext{coth}^{-1}(\sqrt{t}) ight). $$Use the chain rule: first find the derivative of $$	ext{coth}^{-1}(x)$$ with respect to $$x$$, then multiply by the derivative of $$\sqrt{t}$$ with respect to $$t. $$Recall that $$\frac{d}{dx} 	ext{coth}^{-1}(x) = -\frac{1}{1 - x^2$}$$ for $$|x| \u003e 1. $$Then, $$\frac{d}{dt} \sqrt{t} = \frac{1}{2\sqrt{t}}. $$Combine these to find $$v' = -\frac{1}{1 - (\sqrt{t})^2$} \cdot \frac{1}{2\sqrt{t}} = -\frac{1}{1 - t} \cdot \frac{1}{2\sqrt{t}}. $$Substitute $$u$$, $$u'$$, $$v$$, and $$v'$$ back into the product rule formula to express $$\frac{dy}{dt} = (-1) \cdot 	ext{coth}^{-1}(\sqrt{t}) + (1 - t) \cdot \left(-\frac{1}{1 - t} \cdot \frac{1}{2\sqrt{t}}ight). $$Simplify the expression as much as possible.

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
29. y = (1 - t)coth⁻¹(√t)

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Inverse Hyperbolic Cotangent Function (coth⁻¹)

Chain Rule

Product Rule