Find \(\frac{\mathrm{d} y}{\mathrm{~d} t}\) where \(y\) is given by (a) \(\mathrm{e}^{\left(3 r^{2}\right)}\) (b) \(3 \mathrm{e}^{\mathrm{t}^{-1}}\) (c) \(\mathrm{e}^{\sin 2 t}\) (d) \(\mathrm{e}^{2 \sin t}\) (e) \(2 e^{\sin t}\)

Differentiating exponential functions, especially those that involve the natural base e, is a critical concept in calculus. The basic rule is that the derivative of e to the power of a function is the original exponential function times the derivative of the exponent. For example, if you have y = e^x, then the derivative dy/dx equals e^x. This rule remains true for any differentiable function in the exponent, such as y = e^(3r2), which was encountered in part (a) of the exercise.

After differentiating the exponent, we obtain the derivative of the entire expression by multiplying the original function by this derivative. This leads to dy/dr = 6re^(3r2) in our example. The concept applies in all similar situations, so you should focus on first finding the derivative of the power function, then multiply it by the original exponential function to get the final derivative.

The chain rule is a fundamental tool for finding the derivative of composite functions, such as y = e^f(x) where f(x) is another function of x. It tells us to take the derivative of the outer function, pretending the inner function is a constant (which we call 'freezing the inside'), and then multiply that result by the derivative of the inner function.

The steps in the given solutions illustrate the chain rule in action. For instance, with y = 3e^t-1, the outer function is e^x and the inner function is t^-1. By applying the chain rule, we can find the derivative dy/dt by multiplying the derivative of the outer function, 3e^t-1, by the derivative of t^-1, which is -t^-2. This results in -3t^-2e^t-1. Understanding the chain rule is essential for handling more complex differentiation problems involving multiple functions nested within each other.

When it comes to differentiating trigonometric functions as part of an exponential function, as in parts (c) to (e) of the exercise, we must apply the rules for differentiating both exponential and trigonometric functions, often using the chain rule. The derivatives of the six basic trigonometric functions are well established. For example, the derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x).

In part (c), the derivative of y = e^sin(2t) is found by first differentiating the outer function (the exponential function) normally, retaining sin(2t) as the exponent, and then multiplying by the derivative of sin(2t), which is 2cos(2t). This application of the chain rule simplifies the differentiation process, allowing us to handle even complex combinations of functions, such as those with trigonometric expressions within an exponential function.