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Chapter 3: Problem 17

Find the following derivatives. $$\frac{d}{d x}\left(\left(x^{2}+1\right) \ln x\right)$$

Short Answer

Expert verified
Question: Find the derivative of the function \((x^2+1)\ln x\). Answer: The derivative of the given function is: $$\frac{d}{d x}\left(\left(x^{2}+1\right) \ln x\right)= 2x\ln x + x + 1 - \frac{1}{x}$$

Step by step solution

01

Find the derivative of u(x)

In this case, \(u(x)=x^2+1\). To find its derivative, \(\frac{d}{d x}(x^2+1)\), we can use basic differentiation rules: $$u'(x) = \frac{d}{d x}(x^2+1) = 2x$$
02

Find the derivative of v(x)

In this case, \(v(x)=\ln x\). To find its derivative, \(\frac{d}{d x}(\ln x)\), we can use basic differentiation rules: $$v'(x) = \frac{d}{d x}(\ln x) = \frac{1}{x}$$
03

Apply the Product Rule formula

Now we have both \(u'(x)\) and \(v'(x)\). We will apply the Product Rule formula to find the derivative of the given function: $$\frac{d}{d x}((x^2+1)\ln x) = u'(x)v(x) + u(x)v'(x) = 2x\ln x + (x^2+1)\left(\frac{1}{x}\right)$$
04

Simplify the expression

Let's simplify the expression to get the final answer: $$\frac{d}{d x}((x^2+1)\ln x) = 2x\ln x + x + 1 - \frac{1}{x}$$ Therefore, the derivative of the given function is: $$\frac{d}{d x}\left(\left(x^{2}+1\right) \ln x\right)= 2x\ln x + x + 1 - \frac{1}{x}$$

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