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Find the derivative of $y$ . $\newline$ $y=3t^{3}\tanh\left(\frac{1}{t^{2}}\right)$

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Find derivatives of inverse trigonometric functions

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Q. Find the derivative of

y

.

\newline

y=3t^{3}\tanh\left(\frac{1}{t^{2}}\right)

Apply Product Rule: step_1: Apply the product rule for differentiation, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. $\newline$ Let $u = 3t^3$ and $v = \tanh(\frac{1}{t^2})$ . Then $y = u \cdot v$ . $\newline$ We need to find $u'$ (the derivative of $u$ with respect to $t$ ) and $v'$ (the derivative of $v$ with respect to $t$ ).
Differentiate $u$ : step_2: Differentiate $u = 3t^3$ with respect to $t$ .
$u' = \frac{d}{dt}(3t^3) = 3 \cdot \frac{d}{dt}(t^3) = 3 \cdot 3t^2 = 9t^2.$
Differentiate $v$ : step_3: Differentiate $v = \tanh(\frac{1}{t^2})$ with respect to $t$ using the chain rule. $\newline$ The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. $\newline$ v' = \frac{d}{dt}(\tanh(\frac{1}{t^2})) = \sech^2(\frac{1}{t^2}) \cdot \frac{d}{dt}(\frac{1}{t^2}).
Differentiate Inner Function: step_4: Differentiate the inner function $\frac{1}{t^2}$ with respect to $t$ . Let $w = \frac{1}{t^2}$ , then $w' = \frac{d}{dt}(\frac{1}{t^2}) = \frac{d}{dt}(t^{-2}) = -2t^{-3} = -\frac{2}{t^3}$ .
Combine Results: step_5: Combine the results from step $3$ and step $4$ to find $v'$ . $v' = \text{sech}^2(\frac{1}{t^2}) \cdot \left(-\frac{2}{t^3}\right)$ .
Use Product Rule: step_6: Use the product rule to find the derivative of $y = u \cdot v$ . $y' = u' \cdot v + u \cdot v'$ .Substitute $u' = 9t^2$ from step $2$ and $v'$ from step $5$ into the equation.y' = 9t^2 \cdot \tanh(\frac{1}{t^2}) + 3t^3 \cdot (\sech^2(\frac{1}{t^2}) \cdot (-\frac{2}{t^3})).
Simplify Expression: step_7: Simplify the expression for $y'$ .y' = 9t^2 \cdot \tanh(\frac{1}{t^2}) - 6 \cdot \sech^2(\frac{1}{t^2}).
Simplify Expression: step extunderscore $7$ : Simplify the expression for $y'$ . y' = 9t^2 \cdot \tanh(\frac{1}{t^2}) - 6 \cdot \sech^2(\frac{1}{t^2}). $\text{total\textunderscore number\textunderscore of\textunderscore steps}=7$

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