Find the 2nd Derivative y=x natural log of x

$y = x \ln (x)$

Find the first derivative.

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Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (x)$ .

$f' (x) = x \frac{d}{d x} (\ln (x)) + \ln (x) \frac{d}{d x} (x)$

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$f' (x) = x (\frac{1}{x}) + \ln (x) \frac{d}{d x} (x)$

Differentiate using the Power Rule.

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Combine $x$ and $\frac{1}{x}$ .

$f' (x) = \frac{x}{x} + \ln (x) \frac{d}{d x} (x)$

Cancel the common factor of $x$ .

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Cancel the common factor.

$f' (x) = \frac{x}{x} + \ln (x) \frac{d}{d x} (x)$

Divide $1$ by $1$ .

$f' (x) = 1 + \ln (x) \frac{d}{d x} (x)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = 1 + \ln (x) \cdot 1$

Multiply $\ln (x)$ by $1$ .

$f' (x) = 1 + \ln (x)$

Find the second derivative.

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Differentiate.

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By the Sum Rule, the derivative of $1 + \ln (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\ln (x)]$ .

$f'' (x) = \frac{d}{d x} (1) + \frac{d}{d x} (\ln (x))$

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (\ln (x))$

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$f'' (x) = 0 + \frac{1}{x}$

Add $0$ and $\frac{1}{x}$ .

$f'' (x) = \frac{1}{x}$

Find the third derivative.

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Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$f''' (x) = \frac{d}{d x} (x^{- 1})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 1$ .

$f''' (x) = - x^{- 2}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f''' (x) = - \frac{1}{x^{2}}$

Find the fourth derivative.

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Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{x^{2}}$ .

$f^{4} (x) = - \frac{d}{d x} \cdot \frac{1}{x^{2}} + \frac{1}{x^{2}} \cdot \frac{d}{d x} (- 1)$

Differentiate.

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Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$f^{4} (x) = - \frac{d}{d x} {(x^{2})}^{- 1} + \frac{1}{x^{2}} \cdot \frac{d}{d x} (- 1)$

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{4} (x) = - \frac{d}{d x} x^{2 \cdot - 1} + \frac{1}{x^{2}} \cdot \frac{d}{d x} (- 1)$

Multiply $2$ by $- 1$ .

$f^{4} (x) = - \frac{d}{d x} x^{- 2} + \frac{1}{x^{2}} \cdot \frac{d}{d x} (- 1)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 2$ .

$f^{4} (x) = - (- 2 x^{- 3}) + \frac{1}{x^{2}} \cdot \frac{d}{d x} (- 1)$

Multiply $- 2$ by $- 1$ .

$f^{4} (x) = 2 x^{- 3} + \frac{1}{x^{2}} \cdot \frac{d}{d x} (- 1)$

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f^{4} (x) = 2 x^{- 3} + \frac{1}{x^{2}} \cdot 0$

Simplify the expression.

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Multiply $\frac{1}{x^{2}}$ by $0$ .

$f^{4} (x) = 2 x^{- 3} + 0$

Add $2 x^{- 3}$ and $0$ .

$f^{4} (x) = 2 x^{- 3}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f^{4} (x) = 2 (\frac{1}{x^{3}})$

Combine $2$ and $\frac{1}{x^{3}}$ .

$f^{4} (x) = \frac{2}{x^{3}}$

Find the 2nd Derivative y=x natural log of x

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