Use Logarithmic Differentiation to Find the Derivative y=x^(2x)

$y = x^{2 x}$

Let $y = f (x)$ , take the natural logarithm of both sides $\ln (y) = \ln (f (x))$ .

$\ln (y) = \ln (x^{2 x})$

Expand $\ln (x^{2 x})$ by moving $2 x$ outside the logarithm.

$\ln (y) = 2 x \ln (x)$

Differentiate the expression using the chain rule, keeping in mind that $y$ is a function of $x$ .

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Differentiate the left hand side $\ln (y)$ using the chain rule.

$\frac{y'}{y} = 2 x \ln (x)$

Differentiate the right hand side.

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Differentiate $2 x \ln (x)$ .

$\frac{y'}{y} = \frac{d}{d x} [2 x \ln (x)]$

Since $2$ is constant with respect to $x$ , the derivative of $2 x \ln (x)$ with respect to $x$ is $2 \frac{d}{d x} [x \ln (x)]$ .

$\frac{y'}{y} = 2 \frac{d}{d x} [x \ln (x)]$

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 $g (x) = \ln (x)$ .

$\frac{y'}{y} = 2 (x \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x])$

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

$\frac{y'}{y} = 2 (x \frac{1}{x} + \ln (x) \frac{d}{d x} [x])$

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

$\frac{y'}{y} = 2 (\frac{x}{x} + \ln (x) \frac{d}{d x} [x])$

Cancel the common factor of $x$ .

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Cancel the common factor.

$\frac{y'}{y} = 2 (\frac{x}{x} + \ln (x) \frac{d}{d x} [x])$

Rewrite the expression.

$\frac{y'}{y} = 2 (1 + \ln (x) \frac{d}{d x} [x])$

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

$\frac{y'}{y} = 2 (1 + \ln (x) \cdot 1)$

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

$\frac{y'}{y} = 2 (1 + \ln (x))$

Simplify.

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Apply the distributive property.

$\frac{y'}{y} = 2 \cdot 1 + 2 \ln (x)$

Multiply $2$ by $1$ .

$\frac{y'}{y} = 2 + 2 \ln (x)$

Reorder terms.

$\frac{y'}{y} = 2 \ln (x) + 2$

Isolate $y'$ and substitute the original function for $y$ in the right hand side.

$y' = (2 \ln (x) + 2) x^{2 x}$

Simplify the right hand side.

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Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

$y' = (\ln (x^{2}) + 2) x^{2 x}$

Apply the distributive property.

$y' = \ln (x^{2}) x^{2 x} + 2 x^{2 x}$

Reorder factors in $\ln (x^{2}) x^{2 x} + 2 x^{2 x}$ .

$y' = x^{2 x} \ln (x^{2}) + 2 x^{2 x}$

Use Logarithmic Differentiation to Find the Derivative y=x^(2x)

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