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Ross' question via email about a derivative.

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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.
 
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