## Related questions with answers

Question

Find the first derivatives of the functions.

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

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 2In this exercise we have:

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

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