Try the fastest way to create flashcards
Question

Find the first derivatives of the functions.

y=sin[(2t+5)2/3]y=\sin [(2 t+5)^{-2 / 3}]

Solution

Verified
Answered 2 years ago
Answered 2 years ago
Step 1
1 of 2

In this exercise we have:

 ⁣dy ⁣dt=d ⁣dt(sin[(2t+5)2/3])=cos[(2t+5)2/3]Chain Ruled ⁣dt((2t+5)2/3)=cos[(2t+5)2/3](23)(2t+5)2/31Chain Ruled ⁣dt(2t+5)=43cos[(2t+5)2/3](2t+5)5/3\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*}

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Thomas' Calculus 14th Edition by Christopher E Heil, Joel R. Hass, Maurice D. Weir

Thomas' Calculus

14th EditionISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir
10,142 solutions
Thomas' Calculus 11th Edition by Frank R. Giordano, George B Thomas Jr, Joel D. Hass, Maurice D. Weir

Thomas' Calculus

11th EditionISBN: 9780321185587Frank R. Giordano, George B Thomas Jr, Joel D. Hass, Maurice D. Weir
10,789 solutions
Calculus: Early Transcendentals 8th Edition by James Stewart

Calculus: Early Transcendentals

8th EditionISBN: 9781285741550James Stewart
11,083 solutions
Calculus: Early Transcendentals 9th Edition by Daniel K. Clegg, James Stewart, Saleem Watson

Calculus: Early Transcendentals

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

More related questions

1/4

1/7