frac{d}{dt}(tan t + t^2 operatorname{cosech} | vedclass

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(B) To find the derivative of $	an t + t^2 \operatorname{cosech} t$ with respect to $t$, we use the sum rule and the product rule. The derivative of

Vedclass · Accepted Answer

$\sec^2 t + 2t \operatorname{cosech} t - t^2 \operatorname{cosech} t \operatorname{coth} t$

Vedclass · Answer

(B) To find the derivative of $	an t + t^2 \operatorname{cosech} t$ with respect to $t$, we use the sum rule and the product rule. The derivative of $	an t$ is $\sec^2 t$. Using the product rule for $t^2 \operatorname{cosech} t$, we have: $\frac{d}{dt}(t^2 \operatorname{cosech} t) = \frac{d}{dt}(t^2) \cdot \operatorname{cosech} t + t^2 \cdot \frac{d}{dt}(\operatorname{cosech} t)$. Since $\frac{d}{dt}(t^2) = 2t$ and $\frac{d}{dt}(\operatorname{cosech} t) = -\operatorname{cosech} t \operatorname{coth} t$, we get: $2t \operatorname{cosech} t + t^2(-\operatorname{cosech} t \operatorname{coth} t) = 2t \operatorname{cosech} t - t^2 \operatorname{cosech} t \operatorname{coth} t$. Combining these, the final result is $\sec^2 t + 2t \operatorname{cosech} t - t^2 \operatorname{cosech} t \operatorname{coth} t$.

$\frac{d}{dt}(\tan t + t^2 \operatorname{cosech} t)$ is equal to

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