What is the derivative of x^{log_{e}(x}) ?

The derivative of x^{log_{e}(x}) is (2e^{ln^2(x)}ln(x))/x

What is the first derivative of x^{log_{e}(x}) ?

The first derivative of x^{log_{e}(x}) is (2e^{ln^2(x)}ln(x))/x

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\frac{d}{d x} (x^{l o g_{e} (x)})

\frac{2 e ^{l n^{2} (x)} ln ( x )}{x}

Solution steps

\frac{d}{d x} (x^{l o g_{e} (x)})

Simplify

x^{l o g_{e} (x)} : x^{l n (x)}

= \frac{d}{d x} (x^{l n (x)})

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{l n (x)} = e^{l n (x) l n (x)}

= \frac{d}{d x} (e^{l n (x) l n (x)})

Apply the chain rule:

e^{l n (x) l n (x)} \frac{d}{d x} (ln (x) ln (x))

= e^{l n (x) l n (x)} \frac{d}{d x} (ln (x) ln (x))

\frac{d}{d x} (ln (x) ln (x)) = \frac{2 ln ( x )}{x}

= e^{l n (x) l n (x)} \frac{2 ln ( x )}{x}

Simplify

e^{l n (x) l n (x)} \frac{2 ln ( x )}{x} : \frac{2 e ^{l n^{2} (x)} ln ( x )}{x}

= \frac{2 e ^{l n^{2} (x)} ln ( x )}{x}

\frac{1}{3 + t}

x \to 3 lim (\frac{x ^{2} - 4}{x - 2})

x + y = \frac{d y}{d x}

\frac{d}{d x} (x ln (x^{9}))

\frac{d}{d x} ((1 + x^{2}) (x^{\frac{3}{4}} - x^{- 3}))

What is the derivative of x^{log_{e}(x}) ?
The derivative of x^{log_{e}(x}) is (2e^{ln^2(x)}ln(x))/x
What is the first derivative of x^{log_{e}(x}) ?
The first derivative of x^{log_{e}(x}) is (2e^{ln^2(x)}ln(x))/x