Find the derivative of y.
-$y = \operatorname { sech } ( 5 \theta ) ( 1 - \ln \operatorname { sech } ( 5 \theta ) )$
A) $1 - 5 \operatorname { sech } ( 5 \theta ) \tanh ( 5 \theta ) \ln \operatorname { sech } ( 5 \theta )$
B) $5 \operatorname { sech } ( 5 \theta ) \tanh ( 5 \theta ) \ln ( 5 \theta )$
C) $\operatorname { sech } ( 5 \theta ) \tanh ( 5 \theta ) \ln \operatorname { sech } ( 5 \theta )$
D) $5 \operatorname { sech } ( 5 \theta ) \tanh ( 5 \theta ) \ln \operatorname { sech } ( 5 \theta )$

Question

Quizplus · Accepted Answer

To find the derivative of $y = \operatorname { sech } ( 5 \theta ) ( 1 - \ln \operatorname { sech } ( 5 \theta ) )$, we apply the product rule and the chain rule. The derivative of $\operatorname { sech } ( 5 \theta )$ is $-5 \operatorname { sech } ( 5 \theta ) \tanh ( 5 \theta )$, and the derivative of $1 - \ln \operatorname { sech } ( 5 \theta )$ is $\frac{5 \tanh(5\theta)}{\operatorname { sech } ( 5 \theta )}$ due to the chain rule. Multiplying these derivatives according to the product rule and simplifying gives the correct answer.

Find the Derivative of Y y=sech⁡(5θ)(1−ln⁡sech⁡(5θ))y = \operatorname { sech } ( 5 \theta ) ( 1 - \ln \operatorname { sech } ( 5 \theta ) )y=sech(5θ)(1−lnsech(5θ))

Find the Derivative of Y $y = \operatorname { sech } ( 5 \theta ) ( 1 - \ln \operatorname { sech } ( 5 \theta ) )$