The slope of the tangent to curve is

JEE MainMathematicsModerateDefinite Integration

The slope of the tangent to curve
$y = \int_{x}^{x^{2}}\mspace{2mu}\cos^{- 1}t^{2}dt\ at\ x = \frac{1}{\sqrt[4]{2}}$ is

Show answer and solution

\frac{dy}{dx} = \cos^{- 1}\left( x^{4} \right) \cdot 2x - \cos^{- 1}\left( x^{2} \right).1

\left. \ \frac{dy}{dx} \right|_{x = \frac{1}{\sqrt{2}}} = \left( \frac{\sqrt[4]{8}}{3} - \frac{1}{4} \right)\pi

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

Scan any doubt from your notes and get a whiteboard explanation.