Exponential Functions and Their Derivatives
Exponential functions are of the form f(x) = ax, where "a" is a positive constant. When a = e (approximately 2.71828), the function becomes particularly significant due to its unique properties when differentiated and integrated. For a broader understanding of exponential functions, see our notes on Exponential Functions.
Derivative of ex
- The derivative of ex with respect to x is itself, i.e., (d/dx)ex = ex.
- This unique property arises due to the limit definition of the derivative and the particular growth rate of ex. To delve into the foundational concepts of differentiation, consider reading about Basic Differentiation Rules.
Example 1: Differentiate f(x) = 3ex
Using the constant multiple rule, we find that f'(x) = 3(d/dx)ex = 3ex.
Example 2: Find the tangent line to y = ex at x = 0
Since (d/dx)ex = ex, the slope at x = 0 is e0 = 1. Using the point-slope form of a line, y - 1 = 1(x - 0), so the tangent line is y = x + 1.
Logarithmic Functions and Their Derivatives
Logarithmic functions are the inverse of exponential functions. The natural logarithm, ln(x), is the inverse of the exponential function ex. A detailed exploration of logarithmic functions can be found in our notes on Logarithmic Functions.
Derivative of ln(x)
- The derivative of ln(x) with respect to x is 1/x.
- This result can be derived using the limit definition of the derivative and properties of logarithms.
Example 3: Differentiate g(x) = ln(5x)
Using the chain rule, g'(x) = (1/(5x)) * 5 = 1/x.
Example 4: Find the slope of the tangent to y = ln(x) at x = 1
Since (d/dx)ln(x) = 1/x, the slope at x = 1 is 1/1 = 1. Thus, the tangent line at x = 1 is 1.
Applications in Real-World Contexts
Exponential Growth and Decay
- Exponential functions are widely used to model growth and decay phenomena, such as population growth, radioactive decay, and interest calculations. The principle of exponential growth can be further understood through our notes on Compound Interest.
- The derivative, ex, helps analyse the rate of change in these contexts, providing insights into the speed and nature of the growth or decay.
Logarithmic Scales
- Logarithmic functions are utilised in creating scales that accommodate a wide range of values, such as the Richter scale for earthquake magnitudes and the pH scale for acidity.
- The derivative, 1/x, is crucial in understanding how small changes in x affect the function, which is vital in scientific analyses and calculations. To see how these functions are integrated, review our notes on Integration of Exponential and Logarithmic Functions.
Exam-Style Questions Embedded in the Notes
Question: Differentiate h(x) = e(2x) and find the slope of the tangent at x = 0
To differentiate h(x) = e(2x), we utilise the chain rule since we have a function within the exponential function. The chain rule states that if we have a composite function h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x). Applying this to e(2x), we get h'(x) = e(2x) * 2 = 2e(2x). To find the slope of the tangent at x = 0, we substitute this value into the derivative: h'(0) = 2e(2*0) = 2e0 = 2. Therefore, the slope of the tangent to the curve at x = 0 is 2.
Question: Differentiate k(x) = ln(x2 + 1) and find the derivative at x = 1
To differentiate k(x) = ln(x2 + 1), we again employ the chain rule. Letting u = x2 + 1, we have k(x) = ln(u). The derivative k'(x) = (1/u) * u'(x) = (1/(x2 + 1)) * (2x). To find the derivative at x = 1, we substitute this value in: k'(1) = (1/(12 + 1)) * (2*1) = 2/2 = 1. Thus, the derivative of k(x) at x = 1 is 1.
FAQ
When differentiating logarithmic functions with bases other than e, the chain rule becomes particularly crucial. Suppose you have a function y = logb(x), where b is a base other than e. To differentiate this, you would use the change of base formula in logarithms, which states logb(a) = ln(a)/ln(b). So, y = ln(x)/ln(b). Now, applying the chain rule, dy/dx = (1/x) * (1/ln(b)), since the derivative of ln(x) with respect to x is 1/x. Thus, the chain rule enables us to differentiate logarithmic functions of any base by converting them to natural logarithms.
The product rule is applied to the differentiation of a product of two functions. If you have a function y = u(x) * v(x), where u and v are functions of x, then dy/dx = u'(x) * v(x) + u(x) * v'(x). In the context of logarithmic functions, suppose y = x * ln(x). Here, u(x) = x and v(x) = ln(x). Differentiating u and v separately, we get u'(x) = 1 and v'(x) = 1/x. Applying the product rule, dy/dx = 1 * ln(x) + x * (1/x) = ln(x) + 1. This application is vital in problems where logarithmic functions are multiplied by algebraic or other transcendental functions.
The functions ex and ln(x) are inverses of each other, meaning that they "undo" each other's operations. Mathematically, e(ln(x)) = x and ln(ex) = x. This inverse relationship is pivotal in solving equations and simplifying expressions in calculus, especially in integrals and derivatives. In mathematical models, particularly those involving growth, decay, or change, ex might be used to model a phenomenon, while ln(x) could be used to linearise exponential relationships or solve for variables in exponential contexts, leveraging their inverse relationship.
The number e, approximately equal to 2.71828, is ubiquitously used in calculus due to its unique properties when differentiated and integrated. When you take the derivative of ex with respect to x, you get ex back, and similarly, the integral of ex dx is also ex. This property is exclusive to the base e and does not hold for other bases in exponential functions. Furthermore, e arises naturally in many mathematical models, especially those describing growth or decay, making it a fundamental base for exponential and logarithmic functions in calculus and applied mathematics.
Yes, the quotient rule can be used for differentiating logarithmic functions when the function is expressed as a quotient. The quotient rule states that if you have a function y = u(x)/v(x), then dy/dx = (v(x) * u'(x) - u(x) * v'(x))/[v(x)]2. For instance, consider y = ln(x)/x. Here, u(x) = ln(x) and v(x) = x. Differentiating u and v, we get u'(x) = 1/x and v'(x) = 1. Applying the quotient rule, dy/dx = (x * (1/x) - ln(x) * 1)/x2 = (1 - ln(x))/x2. Thus, the quotient rule can be applied to differentiate logarithmic functions expressed as quotients, providing a method to explore the rate of change in such functions.
Practice Questions
To differentiate the function y = e(3x) * ln(x), we need to use the product rule since it is a product of two functions of x: u(x) = e(3x) and v(x) = ln(x). The product rule states that if y = u(x) * v(x), then dy/dx = u'(x) * v(x) + u(x) * v'(x).
The derivative of e^(3x) with respect to x is 3e(3x). The derivative of ln(x) with respect to x is 1/x. Applying the product rule, we get: dy/dx = (3e(3x)) * ln(x) + e(3x) * (1/x) dy/dx = 3e(3x) * ln(x) + e(3x)/x
To find the second derivative, we first need to find the first derivative. The derivative of ln(4x) with respect to x can be found using the chain rule for differentiation. Let u(x) = 4x and v(u) = ln(u). Then, y = v(u(x)) and by the chain rule, dy/dx = v'(u) * u'(x).
The derivative of ln(u) with respect to u is 1/u and the derivative of 4x with respect to x is 4. Substituting these into the chain rule, we get: dy/dx = (1/(4x)) * 4 dy/dx = 1/x
Now, to find the second derivative, we differentiate dy/dx = 1/x with respect to x. The derivative of 1/x with respect to x is -1/x2. Therefore, the second derivative of y with respect to x is: d2y/dx2 = -1/x2
