Question

1-22: Differentiate.

22. \(f\left( t \right) = t{e^t}\cot t\)

Short answer

The derivative of the function is \(f'\left( t \right) = {e^t}\left( {\cot t + t\cot t - t{{\csc }^2}t} \right)\).

Step-by-step solution

 Derivative of Trigonometric Functions

\(\begin{aligned}\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\csc x} \right) &= - \csc x\cot x\\\frac{d}{{dx}}\left( {\cos x} \right) &= - \sin x\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\sec x} \right) &= \sec x\tan x\\\frac{d}{{dx}}\left( {\tan x} \right) &= {\sec ^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\cot x} \right) &= - {\csc ^2}x\end{aligned}\)

Differentiate the function

Exercise 3.2.63(a) states that the Product Rule can be extended to the product of three functions is given by \({\left( {fgh} \right)^\prime } = f'gh + fg'h + fgh'\).

Use exercise 3.2.63(a) to differentiate the function as shown below:

\(\begin{aligned}f'\left( t \right) &= \frac{d}{{dt}}\left( {t{e^t}\cot t} \right)\\ &= {e^t}\cot t\frac{d}{{dt}}\left( t \right) + t\cot t\frac{d}{{dt}}\left( {{e^t}} \right) + t{e^t}\frac{d}{{dt}}\left( {\cot t} \right)\\ &= {e^t}\cot t\left( 1 \right) + t\cot t\left( {{e^t}} \right) + t{e^t}\left( { - {{\csc }^2}t} \right)\\ &= {e^t}\cot t + t{e^t}\cot t - t{e^t}{\csc ^2}t\\ &= {e^t}\left( {\cot t + t\cot t - t{{\csc }^2}t} \right)\end{aligned}\)

Thus, the derivative of the function is \(f'\left( t \right) = {e^t}\left( {\cot t + t\cot t - t{{\csc }^2}t} \right)\).