# What is the derivative of #y=x^lnx#?

Take natural log of both sides

Rewrite write hand side using properties of logarithms

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To find the derivative of ( y = x^{\ln x} ), you can use logarithmic differentiation. Taking the natural logarithm of both sides, we get:

[ \ln y = \ln(x^{\ln x}) ] [ \ln y = \ln x \cdot \ln x ] [ \ln y = (\ln x)^2 ]

Now, differentiate implicitly with respect to ( x ):

[ \frac{1}{y} \cdot \frac{dy}{dx} = 2 \ln x \cdot \frac{1}{x} ] [ \frac{dy}{dx} = y \cdot 2 \ln x \cdot \frac{1}{x} ]

Substituting back the original expression for ( y ), which is ( x^{\ln x} ), we get:

[ \frac{dy}{dx} = x^{\ln x} \cdot 2 \ln x \cdot \frac{1}{x} ] [ \frac{dy}{dx} = 2x^{\ln x - 1} \ln x ]

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To find the derivative of (y = x^{\ln(x)}), use logarithmic differentiation.

[y = x^{\ln(x)}] [\ln(y) = \ln\left(x^{\ln(x)}\right)] [\ln(y) = \ln(x) \cdot \ln(x)] [\ln(y) = (\ln(x))^2]

Now differentiate both sides with respect to (x):

[\frac{1}{y} \cdot \frac{dy}{dx} = 2\ln(x) \cdot \frac{1}{x}] [\frac{dy}{dx} = 2\ln(x) \cdot \frac{1}{x} \cdot y]

Substitute back (y = x^{\ln(x)}):

[\frac{dy}{dx} = 2\ln(x) \cdot \frac{1}{x} \cdot x^{\ln(x)}] [\frac{dy}{dx} = 2x^{\ln(x) - 1}]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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