Definite Differential 1

$\begin{aligned} \cos^{-1}\left(\dfrac{2x}{1+x^2}\right) &= \dfrac{\pi}{2} -\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)\\&=\dfrac{\pi}{2}-2\tan^{-1}x\\\dfrac{d}{dx}\left[ \dfrac{\pi}{2}-2\tan^{-1}x\right] &= -\dfrac{2}{1+{x}^2}\\\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ \dfrac{\pi}{2}-2\tan^{-1}x\right]_{ x = \dfrac{1}{4}}&\large\color{#3D99F6}{=\boxed{-\dfrac{32}{17}}}\end{aligned}$

Let $y = \cos^{-1} \left(\dfrac{2x}{1+x^2} \right)$ . $\implies \dfrac{d}{dx} \cos^{-1} \left(\dfrac{2x}{1+x^2} \right) = \dfrac{dy}{dx}$ and:

$\begin{aligned} \implies \cos y & = \frac{2x}{1+x^2} \quad \quad \small \color{#3D99F6}{\text{Differentiate both side.}} \\ - \color{#3D99F6}{\sin y} \ \frac{dy}{dx} \ \bigg|_{x=\frac{1}{4}} & = \left( \frac{2}{1+x^2} - \frac{4x^2}{(1+x^2)^2} \right) \bigg|_{x=\frac{1}{4}} \quad \quad \small \color{#3D99F6}{\cos y \ \bigg|_{x=\frac{1}{4}} = \frac{2\left(\frac{1}{4}\right)}{1+\left(\frac{1}{4}\right)^2} = \frac{8}{17} \implies \sin y = \frac{15}{17}} \\ \implies \frac{dy}{dx} \ \bigg|_{x=\frac{1}{4}} & = -\color{#3D99F6}{\frac{17}{15}} \left( \frac{32}{17} - \frac{64}{289} \right) = -\frac{32}{17} \approx \boxed{-1.882352941} \end{aligned}$

Using:

$\dfrac{d}{dx} \left( \arccos x \right) = \dfrac{-1}{\sqrt{1-x^2}}$

$\dfrac{d}{dx} \left( f(g(x)) \right) = f'(g(x)) g'(x)$

$\dfrac{d}{dx} \left( \dfrac{f(x)}{g(x)}\right) = \dfrac{f'(x)g(x) - g'(x)f(x)}{[g(x)]^2}$

$\dfrac{d}{dx} \left( \arccos \left( \dfrac{2x}{1+x^2}\right) \right) = \dfrac{-1}{\sqrt{1- \left( \dfrac{2x}{1+x^2} \right)^2}} \cdot \left[ \dfrac{2(1+x^2) - (2x)^2}{(1+x^2)^2}\right]$

Upon plugging in $x = \dfrac{1}{4}$ , I got $-\dfrac{32}{17} \approx \boxed{-1.882}$ .

this problem is too highly rated..

Put x= tan@ it reduces to ( -2 / 1 + x^2) after differentiating apply x=1/4 will lead to -1.88.