Find the derivative of y.
-$$y = - 4 t ^ { 3 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right)$$
A) $- 12 t ^ { 2 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right) + 8 \operatorname { sech } ^ { 2 } \left( \frac { 1 } { t ^ { 2 } } \right)$
B) $- 12 t ^ { 2 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right) + 8 \operatorname { sech } \left( \frac { 1 } { t ^ { 2 } } \right)$
C) $- 12 \mathrm { t } ^ { 2 } \tanh \left( \frac { 1 } { \mathrm { t } ^ { 2 } } \right) + 4 \operatorname { sech } ^ { 2 } \left( \frac { 1 } { \mathrm { t } ^ { 2 } } \right)$
D) $- 12 t ^ { 2 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right) - 8 \operatorname { sech } ^ { 2 } \left( \frac { 1 } { t ^ { 2 } } \right)$

Question

Quizplus · Accepted Answer

The product rule is used to differentiate the given function. The product rule states that if $f(x) = g(x)h(x)$, then $f'(x) = g'(x)h(x) + g(x)h'(x)$. In this case, $g(x) = -4t^3$ and $h(x) = 	anh(\frac{1}{t^2})$. So, $g'(x) = -12t^2$ and $h'(x) = -\frac{2}{t^3}\operatorname{sech}^2(\frac{1}{t^2})$. Therefore,$$y' = -12t^2	anh(\frac{1}{t^2}) + (-4t^3)\left(-\frac{2}{t^3}\operatorname{sech}^2(\frac{1}{t^2})ight) = -12t^2	anh(\frac{1}{t^2}) + 8\operatorname{sech}^2(\frac{1}{t^2}).$$

Find the Derivative of Y y=−4t3tanh⁡(1t2)y = - 4 t ^ { 3 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right)y=−4t3tanh(t21​)

Find the Derivative of Y $y = - 4 t ^ { 3 } \tanh \left( \frac { 1 } { t ^ { 2 } } \right)$