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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
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7장, 문제 7.6.130

130. Use the identity arccot(u)=π/2 - arctan(u) to derive the formula for the derivative of arccot(u) in Table 7.4 from the formula for the derivative of arctan(u).

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Recall the given identity: \(\arccot(u) = \frac{\pi}{2} - \arctan(u)\).
Differentiate both sides of the equation with respect to \(x\). Since \(\frac{\pi}{2}\) is a constant, its derivative is zero, so we have: \(\frac{d}{dx} [\arccot(u)] = \frac{d}{dx} \left[ \frac{\pi}{2} - \arctan(u) \right] = - \frac{d}{dx} [\arctan(u)]\).
Use the chain rule to differentiate \(\arctan(u)\): \(\frac{d}{dx} [\arctan(u)] = \frac{1}{1 + u^2} \cdot \frac{du}{dx}\).
Substitute this result back into the derivative of \(\arccot(u)\): \(\frac{d}{dx} [\arccot(u)] = - \frac{1}{1 + u^2} \cdot \frac{du}{dx}\).
Thus, the derivative formula for \(\arccot(u)\) is derived as \(\frac{d}{dx} [\arccot(u)] = - \frac{1}{1 + u^2} \cdot \frac{du}{dx}\), matching the formula in Table 7.4.

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주요 개념

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