Question

Calculate the derivative of the following functions (i) using the fact that \(b^{x}=e^{x \ln b}\) and (ii) by using logarithmic differentiation. Verify that both answers are the same. $$y=\left(x^{2}+1\right)^{x}$$

Short answer

Question: Find the derivative of the function \(y = (x^2 + 1)^x\) using two different methods: \(b^x = e^{x \ln b}\) and logarithmic differentiation. Answer: Both methods result in the same derivative: $$\frac{dy}{dx} = (x^2 + 1)^x \left(\ln (x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}\right)$$

Step-by-step solution

Convert the given function to exponential form

Rewrite the given function using the provided fact: $$y = (x^2 + 1)^x = e^{x \ln (x^2 + 1)}$$

Differentiate with respect to x using the chain rule

Now we need to find the derivative of the function with respect to x. Using the chain rule, we get: $$\frac{dy}{dx} = e^{x \ln (x^2 + 1)} \cdot \frac{d}{dx}(x \ln (x^2 + 1))$$

Find the derivative of the exponent

Next, we need to find the derivative of the exponent with respect to x: $$\frac{d}{dx}(x \ln (x^2 + 1)) = \ln (x^2 + 1) + x \cdot \frac{d}{dx}(\ln (x^2 + 1))$$

Find the derivative of the logarithm

Now find the derivative of \(\ln (x^2 + 1)\) with respect to x using the chain rule: $$\frac{d}{dx}(\ln (x^2 + 1)) = \frac{d}{dx}(x^2 + 1) \cdot \frac{1}{x^2 + 1} = \frac{2x}{x^2 + 1}$$

Find the final derivative using method 1

Now plug the derivative of the exponent into Step 2: $$\frac{dy}{dx} = e^{x \ln (x^2 + 1)} \left(\ln (x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}\right)$$ Next, we need to find the derivative using logarithmic differentiation.

Apply the natural logarithm to both sides

Take the natural logarithm of both sides of the given equation: $$\ln y = \ln \left((x^2 + 1)^x\right)$$

Use logarithmic properties to simplify the equation

Using the logarithmic property \(\ln (a^b) = b \ln a\), we can rewrite the equation as: $$\ln y = x \ln (x^2 + 1)$$

Differentiate both sides with respect to x

Now we will differentiate both sides of the equation with respect to x: $$\frac{1}{y} \cdot \frac{dy}{dx} = \ln (x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}$$

Multiply both sides by y

To solve for \(\frac{dy}{dx}\), multiply both sides of the equation by y: $$\frac{dy}{dx} = y \left(\ln (x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}\right)$$

Substitute the original equation for y

Now substitute the original equation, \((x^2 + 1)^x\), for y: $$\frac{dy}{dx} = (x^2 + 1)^x \left(\ln (x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}\right)$$ Comparing the results from both methods, we see that the derivatives are the same: $$\frac{dy}{dx} = e^{x \ln (x^2 + 1)} \left(\ln (x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}\right) = (x^2 + 1)^x \left(\ln (x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}\right)$$

Key concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus for computing the derivative of a composite function. It's like peeling an onion, layer by layer, to see how changes in one variable affect the whole. Imagine you have a function that is made up of other functions, the chain rule allows you to differentiate this compound function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

For example, if you have a function such as \( f(g(x)) \), the chain rule tells us that the derivative \( f'(g(x)) \times g'(x) \). In our exercise, \( y = (x^2 + 1)^x \) can be seen as a composition of \( e^{x \ln (x^2 + 1)} \) and \( x \ln (x^2 + 1) \), where \( e^{x \ln (x^2 + 1)} \) is the outer function and \( x \ln (x^2 + 1) \) is the inner one. By applying the chain rule, the derivative of \( y \) is then found by taking the derivative of these nested functions in sequence.

Logarithmic Differentiation Explained

Logarithmic differentiation is a technique that simplifies the process of finding the derivatives of more complicated functions, especially those involving products, quotients, or powers where the exponent is a function of x. The magic of logarithmic differentiation lies in the properties of logarithms that turn multiplication into addition, division into subtraction, and powers into products.

Here's how it works: By taking the natural logarithm \( \ln \) of both sides of an equation like \( y = (x^2 + 1)^x \) and using properties of logarithms, we can transform the difficult-to-differentiate form into a more manageable one. After differentiation, we can solve for the derivative \( \frac{dy}{dx} \) by simply multiplying both sides by \( y \) and substituting back for \( y \) using the original equation. This technique reveals the underpinning structure of the function, which makes calculating the derivative a smoother process.

Exponential Functions and Their Derivatives

Exponential functions are the bread and butter of growth and decay processes in mathematics. They are expressed in the form \( b^x \), where \( b \) is a positive constant. Uniquely, the rate of change of an exponential function is proportional to its current value, which means they have the fascinating property that their derivative is quite similar to the function itself.

To differentiate expressions like \( b^x \) where \( b \) is not the natural base \( e \), we first use the transformation \( b^{x} = e^{x \ln b} \). Then, we can differentiate using our known rules for \( e^{x} \) and chain rule. In the exercise, transforming \( (x^2 + 1)^x \) using the natural base \( e \) allows us to apply the chain rule to find its derivative efficiently.