The slope of the tangent to the curve $y=\int\limits_x^{x^2} \cos ^{-1} t^2 d t \text { at } x=2^{-1 / 4}$, is 

$\left(\frac{\sqrt[4]{8}}{3}-\frac{1}{4}\right) \pi$

We have,

$y=\int\limits_x^{x^2} \cos ^{-1} t^2 d t$

$\Rightarrow \frac{d y}{d x}=2 x \cos ^{-1} x^4-\cos ^{-1} x^2$        [By Leibnitz's rule]

$\Rightarrow \left(\frac{d y}{d x}\right)_{x=2^{-1 / 4}}=2 \times 2^{-1 / 4} \cos ^{-1} \frac{1}{2}-\cos ^{-1} \frac{1}{\sqrt{2}}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{x=2^{-1 / 4}}=2^{3 / 4} \times \frac{\pi}{3}-\frac{\pi}{4}=\left(\frac{\sqrt[4]{8}}{3}-\frac{1}{4}\right) \pi$