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What is the derivative (with respect to t) of $\displaystyle \begin{align*} y = 16\,\left[ \sinh{(7\,t)} \right] ^3 \cosh{(7\,t )} \end{align*}$?

One way to do this is to apply the product rule. To do this, we need to know the derivative of each factor.

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \, \left\{ \left[ \sinh{(7\,t)} \right] ^3 \right\} &= 7 \cdot \cosh{( 7\,t )} \cdot 3\left[ \sinh{(7\,t)} \right] ^2 \\ &= 21\cosh{(7\,t)}\left[ \sinh{(7\,t)} \right] ^2 \end{align*}$

and

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\,\left[ \cosh{(7\,t)} \right] = 7\,\sinh{(7\,t)} \end{align*}$

so that means

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} &= 16 \, \left\{ \left[ \sinh{(7\,t)} \right] ^3 \cdot 7\,\sinh{(7\,t)} + 21\cosh{(7\,t)}\left[ \sinh{(7\,t)} \right] ^2 \cdot \cosh{(7\,t)} \right\} \\ &= 112\,\left[ \sinh{(7\,t)} \right] ^2 \, \left\{ \left[ \sinh{(7\,t)} \right] ^2 + 3\, \left[ \cosh{(7\,t)} \right] ^2 \right\} \end{align*}$A more sophisticated method is to use hyperbolic identities to simplify the function before trying to differentiate.

Since $\displaystyle \begin{align*} \sinh{(2\,x)} \equiv 2\sinh{(x)}\cosh{(x)} \end{align*}$ and $\displaystyle \begin{align*} \cosh{(2\,x)} \equiv 1 + 2\,\left[ \sinh{(x)} \right] ^2 \end{align*}$ that means

$\displaystyle \begin{align*} y &= 16\,\left[ \sinh{(7\,t)} \right] ^3\cosh{(7\,t)} \\ &= 8 \,\left[ \sinh{(7\,t)} \right] ^2 \cdot 2\sinh{(7\,t)}\cosh{(7\,t)} \\ &= 8 \cdot \frac{1}{2} \, \left[ \cosh{(14\,t)} - 1 \right] \sinh{(14\,t)} \\ &= 4\cosh{(14\,t)}\sinh{(14\,t)} - 4\sinh{(14\,t)} \\ &= 2\cdot 2\cosh{(14\,t)}\sinh{(14\,t)} - 4\sinh{(14\,t)} \\ &= 2 \sinh{(28\,t)} - 4\sinh{(14\,t)} \\ \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 2 \cdot 28 \cosh{(28\,t)} - 4\cdot 14\cosh{(14\,t)} \\ &= 56\cosh{(28\,t)} - 56\cosh{(14\,t)} \end{align*}$

This can be shown to be equivalent to the answer given above.

One way to do this is to apply the product rule. To do this, we need to know the derivative of each factor.

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \, \left\{ \left[ \sinh{(7\,t)} \right] ^3 \right\} &= 7 \cdot \cosh{( 7\,t )} \cdot 3\left[ \sinh{(7\,t)} \right] ^2 \\ &= 21\cosh{(7\,t)}\left[ \sinh{(7\,t)} \right] ^2 \end{align*}$

and

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\,\left[ \cosh{(7\,t)} \right] = 7\,\sinh{(7\,t)} \end{align*}$

so that means

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} &= 16 \, \left\{ \left[ \sinh{(7\,t)} \right] ^3 \cdot 7\,\sinh{(7\,t)} + 21\cosh{(7\,t)}\left[ \sinh{(7\,t)} \right] ^2 \cdot \cosh{(7\,t)} \right\} \\ &= 112\,\left[ \sinh{(7\,t)} \right] ^2 \, \left\{ \left[ \sinh{(7\,t)} \right] ^2 + 3\, \left[ \cosh{(7\,t)} \right] ^2 \right\} \end{align*}$A more sophisticated method is to use hyperbolic identities to simplify the function before trying to differentiate.

Since $\displaystyle \begin{align*} \sinh{(2\,x)} \equiv 2\sinh{(x)}\cosh{(x)} \end{align*}$ and $\displaystyle \begin{align*} \cosh{(2\,x)} \equiv 1 + 2\,\left[ \sinh{(x)} \right] ^2 \end{align*}$ that means

$\displaystyle \begin{align*} y &= 16\,\left[ \sinh{(7\,t)} \right] ^3\cosh{(7\,t)} \\ &= 8 \,\left[ \sinh{(7\,t)} \right] ^2 \cdot 2\sinh{(7\,t)}\cosh{(7\,t)} \\ &= 8 \cdot \frac{1}{2} \, \left[ \cosh{(14\,t)} - 1 \right] \sinh{(14\,t)} \\ &= 4\cosh{(14\,t)}\sinh{(14\,t)} - 4\sinh{(14\,t)} \\ &= 2\cdot 2\cosh{(14\,t)}\sinh{(14\,t)} - 4\sinh{(14\,t)} \\ &= 2 \sinh{(28\,t)} - 4\sinh{(14\,t)} \\ \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 2 \cdot 28 \cosh{(28\,t)} - 4\cdot 14\cosh{(14\,t)} \\ &= 56\cosh{(28\,t)} - 56\cosh{(14\,t)} \end{align*}$

This can be shown to be equivalent to the answer given above.

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