Problem 27 - Calculus

Question

Questions \(20-29\) are about the derivative \(d y / d x=c b^{x}\). Find the second derivative of \(y=b^{x}\) and also of \(x=\log _{b} y .\)

Step-by-Step Solution

Answer

The second derivative of \(y = b^x\) is \(b^x(\ln b)^2\); the second derivative of \(x = \log_b y\) is \(-\frac{1}{y (\ln b)^2}\).

Step 1: Differentiate the Function y=b^x

To find the first derivative \( \frac{dy}{dx} \) of the function \( y = b^x \), we apply the formula for the derivative of an exponential function. The derivative of \( y = a^x \) with respect to \( x \) is \( y' = a^x \ln a \). Thus for \( y = b^x \), \( \frac{dy}{dx} = b^x \ln b \).

Step 2: Differentiate the Result Again to Find the Second Derivative

We now differentiate the first derivative \( \frac{dy}{dx} = b^x \ln b \) to find the second derivative \( \frac{d^2y}{dx^2} \). Using the same rule, we get \( \frac{d^2y}{dx^2} = (b^x \ln b) \ln b = b^x (\ln b)^2 \).

Step 3: Express Function as x=log_b(y) and Differentiate

Given \( x=\log_b y \), we rewrite this expression using the definition of a logarithm as \( y = b^x \). We have already found that \( \frac{dy}{dx} = b^x \ln b \). Thus, \( \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{b^x \ln b} \).

Step 4: Differentiate dx/dy Again for Second Derivative of x

Now find the second derivative of \( x \) concerning \( y \), which is \( \frac{d^2x}{dy^2} \). Differentiate \( \frac{dx}{dy} = \frac{1}{b^x \ln b} \) with respect to \( y \). This requires using implicit differentiation and the chain rule. We find that \( \frac{d^2x}{dy^2} = -\frac{(b^x \ln b)(\ln b)}{(b^x \ln b)^3} = -\frac{\ln b}{b^x (\ln b)^2} = -\frac{1}{y(\ln b)^2} \).

Step 5: Conclusion

After computing the derivatives, we conclude: the second derivative of \( y = b^x \) is \( b^x (\ln b)^2 \), and the second derivative of \( x = \log_b y \) is \( -\frac{1}{y (\ln b)^2} \).

Key Concepts

Second Derivative

When we talk about the second derivative of a function, we are exploring the rate at which the first derivative itself is changing. The second derivative can provide information about the concavity of the original function, which helps us understand the behavior of the function in terms of curvature and acceleration.

• The first derivative gives us the slope of the tangent line at any point on the function, or the rate of change of the function.
• The second derivative tells us how the rate of change itself is changing. In simpler terms, it tells us if the function is curving upwards or downwards.

For example, if you have the function \(y = b^x\), the first derivative \(\frac{dy}{dx}\) is \(b^x \ln b\). Taking the derivative of this will give us the second derivative, \(b^x (\ln b)^2\). This indicates that the rate of change of the slope is also proportional to the same base exponential function, multiplied by the logarithm of the base squared.

Exponential Function

An exponential function is a mathematical expression in which a variable appears in the exponent, such as \(y = b^x\). They are characterized by a constant base raised to a variable power, and they grow or decay at rates proportional to their current value.

• The base \(b\) in \(b^x\) determines whether the function is growing (if \(b > 1\)) or decaying (if \(b < 1\)).
• Exponential functions have many important properties such as their unique mechanism of growth. They have applications across sciences, finance, computer science, and more.

Deriving these functions, as seen in the solution, involves using logarithms, specifically the natural logarithm \(\ln b\). The natural logarithm serves as the inverse operation of exponentiation for base \(e\). Additionally, exponential functions often demonstrate smooth and continuous growth and have distinctive graphs that rise quickly.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are complex or involve products and powers that make conventional differentiation cumbersome.

• This method is especially useful for functions wherein a variable is an exponent, creating compound products or quotients.
• It involves taking the natural logarithm of both sides of an equation, allowing the application of properties of logarithms to simplify derivatives.

In the exercise, we used logarithmic differentiation when starting with \(x = \log_b y\) and rewriting this expression as \(y = b^x\). This allowed us to differentiate more smoothly. By applying the rule \(\frac{dy}{dy} = \frac{1}{b^x \ln b}\), a simpler form, we were able to obtain derivatives more fluidly for complex compositions.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions—functions nested within other functions.

• It allows us to differentiate a function that is the combination of two or more layers of functions by breaking it down into its derivative components.
• It takes the form: \(\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\).

In the context of this problem, the chain rule appears when differentiating a function like \(\frac{dx}{dy}\) to find the second derivative by considering the composite and implicit aspects present in the original formulation of \(x = \log_b y\). Essentially, the chain rule provided a method to navigate layered differentiations by approaching each layer individually and multiplying the results, aiding us in the derivative journey.