Evaluate the rate of change of \(H(t)=5 \sin t-3 \cos 2 t\) when (a) \(t=0\) (b) \(t=1.3\).

Understanding how to find the derivative of trigonometric functions is vital for any student of calculus. Trigonometric functions, such as sine and cosine, appear frequently in various areas of mathematics and physics, and knowing how to differentiate them allows one to calculate rates of change and motion.

For instance, in our exercise, the function is given as a combination of sine and cosine functions: \( H(t)=5 \sin t-3 \cos 2 t \). To tackle this, we apply the foundational derivative rules for these functions: the derivative of \( \sin t \) with respect to \( t \) is \( \cos t \), and the derivative of \( \cos t \) is \( -\sin t \). Furthermore, when a trigonometric function has a coefficient or a multiple of \( t \), such as \( \sin(kt) \) or \( a\cos(t) \), we treat the coefficient 'a' as a constant multiplier and apply the chain rule for the multiple 'k'. Using these basic principles, we can systematically compute the derivatives of more complex trigonometric expressions to find the rates of change at any given moment.

The chain rule is an indispensable tool in calculus when dealing with composite functions where one function is nested within another. It allows us to differentiate a function of a function by multiplying the derivative of the outer function by the derivative of the inner function.

In our exercise, we encounter the function \( \cos 2t \), which is a composite function where the inner function is \( 2t \). To find its derivative, we use the chain rule and multiply the derivative of the outer function, which is \( -\sin \), by the derivative of the inner function, which is \( 2 \) in this case. The chain rule might seem complicated at first, but with practice, it becomes a natural step in the process of finding derivatives. By mastering the chain rule, we are equipped to handle a wide array of functions within calculus, opening the door to solving more complex problems.

Evaluating derivatives at specific points, as seen in this exercise, allows us to calculate the exact rate of change of a function at those points. It is directly related to the slope of the tangent line to the function's graph at that point and thus represents how quickly the value of the function is changing there.

After finding the general form of the derivative, \( H'(t) = 5 \cos t + 6 \sin 2t \), evaluating it for specific \( t \) values gives us tangible insight into the behavior of the function \( H(t) \) at those instances. For example, by substituting \( t = 0 \) and \( t = 1.3 \) into the derivative, we can observe that the function is increasing at five units per unit of time at \( t = 0 \) and decreasing at approximately 1.964 units per unit of time at \( t = 1.3 \). This process is at the heart of many applications of calculus, such as understanding motion in physics or the rate of change of quantities in economics.