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Chapter 3: Problem 74

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

Expert verified
Question: Find the derivative of the function \(y = (x^2 + 1)^x\) using two different methods and confirm that the results are the same. Answer: The derivative of the given function is \(\frac{dy}{dx} = e^{x\ln(x^2+1)}\left(x\frac{2x}{x^2+1} + \ln(x^2+1)\right)\).

Step by step solution

01

Rewrite the given function using the property

We can rewrite the function \(y = (x^2 + 1)^x\) as: $$y=e^{x\ln(x^2+1)}$$
02

Apply the chain rule and product rule to differentiate

To find the derivative, we can use the chain rule and the product rule: $$\frac{dy}{dx} = e^{x\ln(x^2+1)}\frac{d(x\ln(x^2+1))}{dx}$$
03

Differentiation of the argument

Now we differentiate the argument \(x\ln(x^2+1)\) with respect to x: $$\frac{d(x\ln(x^2+1))}{dx} = x\frac{d(\ln(x^2+1))}{dx} + \ln(x^2+1)\frac{dx}{dx}$$ Using the chain rule again, we have: $$\frac{d(\ln(x^2+1))}{dx} = \frac{1}{x^2+1}\frac{d(x^2+1)}{dx} = \frac{1}{x^2+1}(2x)$$ Substituting back into our expression, we get: $$\frac{d(x\ln(x^2+1))}{dx} = x\frac{2x}{x^2+1} + \ln(x^2+1)$$ Thus, the derivative of the original function is: $$\frac{dy}{dx} = e^{x\ln(x^2+1)}\left(x\frac{2x}{x^2+1} + \ln(x^2+1)\right)$$ #Method 2: Using logarithmic differentiation#
04

Take the natural logarithm of both sides of the equation

Start by taking the natural logarithm of both sides of the equation: $$\ln y = x\ln(x^2+1)$$
05

Differentiate both sides with respect to x

Now we differentiate both sides with respect to x: $$\frac{1}{y}\frac{dy}{dx} = \frac{d(x\ln(x^2+1))}{dx}$$ We already calculated the derivative of the right side in Method 1: $$\frac{d(x\ln(x^2+1))}{dx} = x\frac{2x}{x^2+1} + \ln(x^2+1)$$
06

Isolate the derivative of the original function y

To find the derivative of the original function, we multiply both sides of the equation with \(y\): $$\frac{dy}{dx} = y\left(x\frac{2x}{x^2+1} + \ln(x^2+1)\right)$$ Since we know that \(y = e^{x\ln(x^2+1)}\), we substitute it into the equation: $$\frac{dy}{dx} = e^{x\ln(x^2+1)}\left(x\frac{2x}{x^2+1} + \ln(x^2+1)\right)$$ Looking at the results from both methods, we confirm that they are the same: $$\frac{dy}{dx} = e^{x\ln(x^2+1)}\left(x\frac{2x}{x^2+1} + \ln(x^2+1)\right)$$

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