# Ross' question via email about a derivative.

• MHB
• Prove It
In summary, the derivative (with respect to t) of the given function is found by using the product rule and simplifying using hyperbolic identities. The final answer is 56 times the hyperbolic cosine of 28t minus 56 times the hyperbolic cosine of 14t.
Prove It
Gold Member
MHB
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.

Last edited:

## 1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of one variable with respect to another. In other words, it measures how much one quantity changes in response to a change in another quantity.

## 2. Why is a derivative important?

Derivatives are important because they allow us to analyze and model the behavior of complex systems. They are used in a wide range of fields, including physics, engineering, economics, and more.

## 3. How is a derivative calculated?

The most common way to calculate a derivative is by using the limit definition: taking the limit of the change in one variable divided by the change in another variable as the change in the second variable approaches zero. This results in the slope of a tangent line to the curve at a specific point.

## 4. What are some real-life applications of derivatives?

Some real-life applications of derivatives include predicting the weather, optimizing financial investments, designing efficient transportation routes, and analyzing the growth of populations.

## 5. Are there different types of derivatives?

Yes, there are several types of derivatives, including the first derivative (also known as the derivative function), second derivative, partial derivative, directional derivative, and more. Each type has its own specific applications and uses in different fields of study.

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