The approximate value of cot ^{-1}(1.001) is | vedclass

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Question

(A) Let $f(x) = \cot^{-1}(x)$. The derivative is $f'(x) = -\frac{1}{1+x^2}$.We use the linear approximation formula: $f(a+h) \approx f(a) + h \cdot f'

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$\frac{\pi}{4}-0.0005$

Vedclass · Answer

(A) Let $f(x) = \cot^{-1}(x)$. The derivative is $f'(x) = -\frac{1}{1+x^2}$.We use the linear approximation formula: $f(a+h) \approx f(a) + h \cdot f'(a)$.Here,let $a = 1$ and $h = 0.001$.Calculate $f(a) = \cot^{-1}(1) = \frac{\pi}{4}$.Calculate $f'(a) = -\frac{1}{1+1^2} = -\frac{1}{2}$.Substitute these values into the formula:$f(1.001) \approx \frac{\pi}{4} + (0.001) \cdot \left(-\frac{1}{2}\right)$.$f(1.001) \approx \frac{\pi}{4} - 0.0005$.

The approximate value of $\cot ^{-1}(1.001)$ is

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