91Ó°ÊÓ

Here we have to find the derivative of an integral with respect to x. So, we will use the Fundamental Theorem of Calculus. According to the Fundamental Theorem of Calculus, if \(F(x) = \int_{a(x)}^{b(x)} f(t) dt\), then \(\frac{d}{dx} F(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)\). Hence, first let's find the derivatives of \(\sqrt{x}\) and \(x^2\) with respect to x: \(a'(x) = \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}}\) and \(b'(x) = \frac{d}{dx} (x^2) = 2x\). Now, we can find the derivative of the integral as: \(\frac{d}{dx} \left(\int_{\sqrt{x}}^{x^{2}} \tan^{-1}\left(\frac{t^{2}}{1+t^{2}}\right) \text { dt }\right) = \tan^{-1}\left(\frac{x^4}{1+x^4}\right) \cdot 2x - \tan^{-1}\left(\frac{x}{1+x}\right) \cdot \frac{1}{2\sqrt{x}}\).

Now, we will find the derivative of the denominator with respect to x: \(\frac{d}{dx} (\sin 2x) = 2 \cos 2x\).

Now, we are ready to use L'Hospital's Rule. By substituting the derivatives in the limit, we get: \(\lim _{x \rightarrow 0} \frac{\tan^{-1}\left(\frac{x^4}{1+x^4}\right) \cdot 2x - \tan^{-1}\left(\frac{x}{1+x}\right) \cdot \frac{1}{2\sqrt{x}}}{2 \cos 2x}\).

As x approaches 0, we can calculate the limit. Remember that \(\tan^{-1}(0) = 0\), and also \(\cos 0 = 1\). So, we have: \(\lim _{x \rightarrow 0} \frac{\tan^{-1}\left(\frac{x^4}{1+x^4}\right) \cdot 2x - \tan^{-1}\left(\frac{x}{1+x}\right) \cdot \frac{1}{2\sqrt{x}}}{2 \cos 2x} = \frac{0 \cdot 0 - 0 \cdot \frac{1}{2\sqrt{0}}}{2 \cdot 1} = \frac{0}{2} = 0\). Therefore, the answer is (A) 0.