Try the fastest way to create flashcards
Question

# Find the first derivatives of the functions.$y=\sin [(2 t+5)^{-2 / 3}]$

Solution

Verified
Step 1
1 of 2

In this exercise we have:

\begin{align*} \dfrac{\dd{y}}{\dd{t}}&=\dfrac{\dd}{\dd{t}}\left(\sin \left[\left(2t+5\right)^{-2/3}\right]\right)\\[5pt] &=\cos \left[\left(2t+5\right)^{-2/3}\right]\overset{\text{Chain Rule}}{\cdot}\dfrac{\dd}{\dd{t}}\left(\left(2t+5\right)^{-2/3}\right)\\[5pt] &=\cos \left[\left(2t+5\right)^{-2/3}\right]\cdot\left(-\dfrac{2}{3}\right)\left(2t+5\right)^{-2/3-1}\overset{\text{Chain Rule}}{\cdot}\dfrac{\dd}{\dd{t}}\left(2t+5\right)\\[5pt] &=-\dfrac{4}{3}\cos \left[\left(2t+5\right)^{-2/3}\right]\cdot\left(2t+5\right)^{-5/3} \end{align*}

## Recommended textbook solutions #### Thomas' Calculus

14th EditionISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir
10,142 solutions #### Thomas' Calculus

11th EditionISBN: 9780321185587Frank R. Giordano, George B Thomas Jr, Joel D. Hass, Maurice D. Weir
10,789 solutions #### Calculus: Early Transcendentals

8th EditionISBN: 9781285741550James Stewart
11,083 solutions #### Calculus: Early Transcendentals

9th EditionISBN: 9781337613927 (2 more)Daniel K. Clegg, James Stewart, Saleem Watson
11,049 solutions