Find the derivative of y.
-$$y = \ln ( \operatorname { sech } ( 4 x + 5 ) )$$
A) $\tanh ( 4 x + 5 )$
B) $- \tanh ( 4 x + 5 )$
C) $- 4 \tanh ( 4 x + 5 )$
D) $\frac { 4 } { \operatorname { sech } ( 4 x + 5 ) }$

Question

Quizplus · Accepted Answer

First, apply the chain rule for logarithmic differentiation:
$$y' = \frac{d}{dx} (\ln u) = \frac{1}{u} \cdot u' $$
where $u = \operatorname{sech} (4x + 5)$. 
Then, apply the chain rule again for the derivative of $\operatorname{sech} (4x + 5)$:
$$y' = \frac{1}{\operatorname{sech} (4x + 5)} \cdot \frac{d}{dx} \operatorname{sech} (4x + 5)$$
$$= \frac{1}{\operatorname{sech} (4x + 5)} \cdot (-4\operatorname{sech} (4x + 5) \tanh (4x + 5))$$
$$= -4 \tanh (4x + 5)$$

Therefore, the answer is $\boxed{\textbf{(C)}}$.

Find the Derivative of Y y=ln⁡(sech⁡(4x+5))y = \ln ( \operatorname { sech } ( 4 x + 5 ) )y=ln(sech(4x+5))

Find the Derivative of Y $y = \ln ( \operatorname { sech } ( 4 x + 5 ) )$