Find the derivative of \( y \) with respect to \( t \) \( y=\sec e^{-1}\left(\frac{1}{8 t^{5}}\right), 0

Ask by Dunn Morrison. in the United States

Mar 09,2025

To find the derivative of \( y \) with respect to \( t \), we first need to rewrite \( y \) as: \[ y = \sec\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \] Using the chain rule: \[ \frac{d y}{d t} = \sec\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \tan\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \cdot \frac{d}{dt}\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \] Next, we need to calculate \(\frac{d}{dt}\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right)\). Let \( z = \frac{1}{8 t^{5}} \), we have: \[ \frac{dz}{dt} = -\frac{5}{8} t^{-6} \] Now we can differentiate using the chain rule: \[ \frac{d}{dt}\left(e^{-1}(z)\right) = \frac{d}{dz}\left(e^{-1}(z)\right) \cdot \frac{dz}{dt} \] Using derivative of \( e^{-1}(z) \): \[ \frac{d}{dz}(e^{-1}(z)) = -\frac{1}{e^{(e^{-1}(z))}} \cdot \frac{dz}{dt} \] Putting it all together: \[ \frac{d y}{d t} = \sec\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \tan\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \left(-\frac{1}{e^{(e^{-1}(\frac{1}{8 t^{5}}))}} \cdot \left(-\frac{5}{8} t^{-6}\right)\right) \] Simplify to find: \[ \frac{d y}{d t} = \sec\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \tan\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \cdot \frac{5}{8e^{(e^{-1}(\frac{1}{8t^{5}}))}} t^{-6} \] So the expression for \( \frac{d y}{d t} \) is: \[ \frac{d y}{d t} = \frac{5}{8} \sec\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \tan\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \cdot \frac{1}{e^{(e^{-1}(\frac{1}{8t^{5}}))}} t^{-6} \] Putting all together, we arrive at the final derivative: \[ \frac{d y}{d t} = \frac{5}{8} \sec\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \tan\left(e^{-1}\left(\frac{1}{8 t^{5}}\right)\right) \cdot \frac{1}{e^{(e^{-1}(\frac{1}{8t^{5}}))}} t^{-6} \]