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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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'

Vedclass · Accepted Answer

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

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

Similar Questions

The approximate value of $x^3-2x^2+3x+2$ at $x=2.01$ is

The approximate value of $\sin(60^{\circ} 0^{\prime} 10^{\prime \prime})$ is (given that $\sqrt{3}=1.732, 1^{\circ}=0.0175^{c}$):

Use differentials to approximate $\sqrt{36.6}$.

Find the approximate value of $f(3.02),$ where $f(x)=3x^{2}+5x+3$.

Find the approximate value of $(32.15)^{\frac{1}{5}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module