Let $f\left(x\right)={\left(\mathrm{s}\mathrm{i}\mathrm{n}?\left({\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?x\right)+\mathrm{s}\mathrm{i}\mathrm{n}?\left({\mathrm{c}\mathrm{o}\mathrm{t}}^{-1}?x\right)\right)}^2-1,\left|x\right|>1$ ${\left(\left({}^{}\right)\left({}^{}\right)\right)}^{}$ . If $\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left({\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?\left(f\right(x\left)\right)\right)$ $\frac{}{}\frac{}{}\frac{}{}\left({}^{}\right)$ and $\text{exception 25:}\frac{}{}$ $y\left(\sqrt[]{3}\right)=\frac{\pi }{6}$ , then $y(-\sqrt[]{3})$ $\text{exception 25:}$ is equal to

<p>Let tan?¹x = θ ⇒ x=tanθ ⇒ sinθ = x/√(1+x²)<br>y = (x/√(1+x²)) + (1/√(1+x²)) = (x+1)/√(1+x²). This is not f(x).<br>Let's follow the solution:<br>y = (x+1)²/(1+x²) - 1 = (2x)/(1+x²) = f(x)<br>Now dy/dx = (1/2√y) * f'(x) = .<br>The solution seems to take y as a different function. Let's assume y = (x/(√(1+x²))) + (1/√(1+x²)) - 1. No.<br>Let's assume y's derivative is taken w.r.t to f(x).<br>y = -tan?¹x + c<br>given y(√3)=π/6 ⇒ π/6 = -π/3 + c ⇒ c=π/2<br>y = cot?¹x. Now y(-√3) = cot?¹(-√3) = 5π/6</p>

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