Find the derivative of y.
-$$y = - 5 t ^ { 3 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right)$$
A) $- 15 t ^ { 2 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right) + 10 \operatorname { sech } \left( \frac { 1 } { t ^ { 2 } } \right)$
B) $- 15 t ^ { 2 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right) + 5 \operatorname { sech } ^ { 2 } \left( \frac { 1 } { t ^ { 2 } } \right)$
C) $- 15 t ^ { 2 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right) - 10 \operatorname { sech } ^ { 2 } \left( \frac { 1 } { t ^ { 2 } } \right)$
D) $- 15 t ^ { 2 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right) + 10 \operatorname { sech } ^ { 2 } \left( \frac { 1 } { t ^ { 2 } } \right)$

Question

Quizplus · Accepted Answer

We can use the chain rule and product rule to find the derivative. Let's start with the derivative of the hyperbolic tangent function:
$$\frac{d}{dx}\tanh\left(\frac{1}{x^2}\right)=-\frac{2}{x^3}\operatorname{sech}^2\left(\frac{1}{x^2}\right)$$
Now, using the chain rule and product rule, we have:
\begin{align*}
\frac{d}{dt}y &= \frac{d}{dt}\left(-5t^3\tanh\left(\frac{1}{t^2}\right)\right)\
&=-15t^2\tanh\left(\frac{1}{t^2}\right) +\frac{d}{dt}\left(-5t^3\right)\cdot\tanh\left(\frac{1}{t^2}\right)\
&=-15t^2\tanh\left(\frac{1}{t^2}\right) + (-15t^2)\cdot\frac{2}{t^3}\operatorname{sech}^2\left(\frac{1}{t^2}\right) \
&= -15t^2\tanh\left(\frac{1}{t^2}\right) + 30\operatorname{sech}^2\left(\frac{1}{t^2}\right)
\end{align*}
Simplifying the second term using the fact that $\operatorname{sech}^2(x) = 1/\cosh^2(x)$, we get the final answer:
$$\frac{d}{dt}y = -15t^2\tanh\left(\frac{1}{t^2}\right) + 10\operatorname{sech}^2\left(\frac{1}{t^2}\right)$$

Find the Derivative of Y y=−5t3tanh⁡(1t2)y = - 5 t ^ { 3 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right)y=−5t3tanh(t21​)

Find the Derivative of Y $y = - 5 t ^ { 3 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right)$