Calculate \(y^{\prime}\) where \(y\) is given by (a) \(\left(t^{2}+1\right) \sin 4 t\) (b) \((3 t+7) \mathrm{e}^{-2 t}\) (c) \(\left(e^{x}+\mathrm{e}^{-2 \mathrm{n}}\right)\left(3 x^{2}-2 x\right)\) (d) \(\sqrt{x} e^{x}\) (e) \(\frac{t^{2}+1}{\mathrm{e}^{t}}\)

Understanding the product rule is crucial for differentiating functions that are the product of two or more functions. When we come across an expression such as \( (t^2 + 1) \sin(4t) \) or \( \sqrt{x} e^x \), we can't simply differentiate each term separately. The product rule allows us to systematically calculate the derivative of a product of functions.

The rule states that for two functions \( u(x) \) and \( v(x) \), their product \( u(x) \cdot v(x) \) has a derivative of \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). To apply the rule effectively:

• Identify the two functions being multiplied (\( u(x) \)) and (\( v(x) \) in the expressions we're differentiating.
• Differentiate each function independently to find \( u'(x) \) and \( v'((x) \).
• Substitute these derivatives back into the formula for the product rule.

For instance, in Step 1, \( u(t) = t^2 + 1 \) and \( v(t) = \sin(4t) \) were differentiated to \( u'(t) = 2t \) and \( v'(t) = 4\cos(4t) \) respectively. The product rule then gave us \( y'(t) = u(t)v'(t) + u'(t)v(t) \) as the final derivative.

The chain rule is another essential derivative tool, especially when dealing with composite functions, where one function is nested within another. Take an example of the derivative of \( \sin(4t) \) with respect to \( t \) -- we have an outer function \( \sin(x) \) and an inner function \( 4t \. The chain rule guides us to first differentiate the outer function, then multiply by the derivative of the inner function.

Expressed mathematically, if \( h(x) = f(g(x)) \) is a composite function, then \( h'(x) = f'(g(x)) \cdot g'(x) \). Here’s the breakdown:

• Differentiate the outer function \( f \) while keeping the inner function \( g \) unchanged.
• Next, differentiate the inner function \( g \) with respect to \( x \).
• Multiply these two derivatives to get the derivative of the composite function, \( h'(x) \).

If we look back at Step 1, \( v'(t) = 4\cos(4t) \) is computed using the chain rule, where the derivative of \( \sin(4t) \) is the cosine of the inner function multiplied by the derivative of the inner function, in this case, \( 4 \. This is a perfect example of the chain rule in action.

When dividing two functions, we can't split the derivative across the numerator and denominator. Instead, we employ the quotient rule. This rule is particularly useful when differentiating a fraction where both the numerator and the denominator are functions of \( x \. Its formula is given by \( \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \) for a quotient \( \frac{u(x)}{v(x)} \.

To use the quotient rule effectively:

• Find the derivatives of both the numerator \( u(x) \) and the denominator \( v(x) \) separately.
• Apply the quotient rule formula, where the derivative of the numerator is multiplied by the original denominator, and then the original numerator is multiplied by the derivative of the denominator.
• Subtract the two terms obtained and divide the result by the square of the original denominator function.

Consider Step 5, which offers a practical example of the quotient rule. The function \( u(t) = t^2 + 1 \) is differentiated to \( u'(t) = 2t \) and \( v(t) = e^t \) is differentiated to \( v'(t) = e^t \). The quotient rule then yields the derivative \( y'(t) \).

Exponential functions often appear in calculus problems. The derivative of an exponential function is unique in that it is proportional to the function itself. For the base \( e \) function, \( f(x) = e^x \), the derivative is \( f'(x) = e^x \.

Here are key points for differentiating exponential functions:

• The base \( e \) exponential function has the same derivative as the function itself.
• For an exponential function with a constant multiplier, such as \( e^{kx} \) where \( k \) is a constant, the derivative will be \( ke^{kx} \) as per the chain rule.
• Even when an exponential function is part of a product or quotient, its differentiation is as per its own rule, further combined with the product or quotient rule as required.

In Step 2, \( v'(t) = -2e^{-2t} \) is obtained by applying the chain rule, which recognizes that the derivative of \( e^{-2t} \) maintains the exponential nature but includes the multiplication by the derivative of the exponent. Similarly, steps involving the derivative of \( e^x \) straightforwardly use this concept of exponential derivative.