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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If $y = \frac{1}{4}u^4$ and $u = \frac{2}{3}x^3 + 5$,then $\frac{dy}{dx} = $

If $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$,then $\frac{dy}{dx} = $

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

If $f(t) = \frac{1 + \operatorname{cosec} t}{1 - \operatorname{cosec} t}$ for $0 < t < \frac{\pi}{2}$ and $f^{\prime}(t) = f(t) g(t)$,then $g(t) =$

The graph of function $f$ contains the points $P(1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( \frac{s^2 + 2s - 3}{s - 1} \right) x - 1 - s$. The value of $f'(1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module