Find the derivative of \( y \) with respect to \( t: y=\sec ^{-1}\left(7 t^{4}-3\right) \). \( \frac{d y}{d t}=\square \) Hint

To find the derivative of \( y = \sec^{-1}(7t^4 - 3) \), we can use the derivative of the inverse secant function, which is given by \[ \frac{d}{dx} \sec^{-1}(x) = \frac{1}{|x| \sqrt{x^2 - 1}}. \] Applying the chain rule, we have: \[ \frac{dy}{dt} = \frac{1}{|7t^4 - 3| \sqrt{(7t^4 - 3)^2 - 1}} \cdot \frac{d}{dt} (7t^4 - 3). \] Calculating the derivative of \( 7t^4 - 3 \): \[ \frac{d}{dt}(7t^4 - 3) = 28t^3. \] Thus, substituting back in gives: \[ \frac{dy}{dt} = \frac{28t^3}{|7t^4 - 3| \sqrt{(7t^4 - 3)^2 - 1}}. \] So the final expression will be: \[ \frac{d y}{d t} = \frac{28t^3}{|7t^4 - 3| \sqrt{(7t^4 - 3)^2 - 1}}. \]