Question

Differentiate the following with respect to the variable: (a) \(y=3 \mathrm{e}^{2 x}\) (b) \(f(t)=\frac{4}{3 \mathrm{e}^{5 t}}\)

Short answer

(a) \(\frac{dy}{dx} = 6e^{2x}\); (b) \(\frac{df}{dt} = -\frac{20}{3} e^{-5t}\).

Step-by-step solution

Differentiate Exponential Function in (a)

To differentiate the function \(y=3e^{2x}\), we use the chain rule. Let \(u=2x\), then \(y=3e^{u}\). The derivative of \(y\) with respect to \(x\) is \(dy/dx = (dy/du) \cdot (du/dx)\). Since \(dy/du = 3e^{u}\) and \(du/dx = 2\), the derivative is: \[\frac{dy}{dx} = 3e^{2x} \cdot 2 = 6e^{2x}.\]

Differentiate Function using Quotient Rule in (b)

To differentiate \(f(t)=\frac{4}{3e^{5t}}\), which can be written as \(f(t) = \frac{4}{3} \cdot e^{-5t}\), we use the chain rule. Let \(v = -5t\), then \(f(t) = \frac{4}{3} e^{v}\). The derivative of \(f\) with respect to \(t\) is \(df/dt = (df/dv) \cdot (dv/dt)\). The derivative of \(f\) concerning \(v\) is \(df/dv = \frac{4}{3} e^{v}\) and \(dv/dt = -5\). Combining, we have: \[\frac{df}{dt} = \frac{4}{3} e^{-5t} \cdot (-5) = -\frac{20}{3} e^{-5t}.\]

Key concepts

Exponential Functions

Exponential functions involve variables in the exponent. They appear frequently in mathematics, science, and real-world applications, such as compound interest and population growth. The general form is

• \(y = a \cdot e^{bx}\)

where \(a\) and \(b\) are constants, and \(e\) is the mathematical constant approximately equal to 2.71828.
To differentiate exponential functions, you usually follow these steps:

• The derivative of \(e^{x}\) is \(e^{x}\) itself.
• For \(e^{u}\), where \(u\) is a function of \(x\), use the chain rule.

This characteristic makes them straightforward to handle in calculus, especially when combined with other rules like the chain rule.

Chain Rule

The chain rule is essential for finding the derivatives of composite functions. This rule applies when you have one function inside another. For a function like \(y = f(g(x))\), the derivative is determined by the formula:

• \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)

where \(u = g(x)\).
This means you first differentiate the outside function with respect to the inside function, and then multiply by the derivative of the inside function. This rule is vital when differentiating expressions where the variable is embedded, such as exponential functions like \(3e^{2x}\). It's like peeling an onion: one layer at a time.

Quotient Rule

The quotient rule is used for finding the derivative of a function that is the division of two other functions. If you have \(f(x) = \frac{g(x)}{h(x)}\), the derivative \(\frac{df}{dx}\) is:

• \(\frac{df}{dx} = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}\)

The numerator involves the derivatives of the numerator and the denominator, while the denominator is simply the square of \(h(x)\).
In some cases, such as with the problem \(\frac{4}{3e^{5t}}\), it can be helpful to express the function differently (e.g., as a product) to simplify differentiation. However, knowing the quotient rule is crucial for cases where the function can't be easily rewritten.