What is the derivative of (3x^{1/x}) ?

The derivative of (3x^{1/x}) is 3^{1/x}x^{(1-2x)/x}(-ln(3x)+1)

What is the first derivative of (3x^{1/x}) ?

The first derivative of (3x^{1/x}) is 3^{1/x}x^{(1-2x)/x}(-ln(3x)+1)

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\frac{d}{d x} ((3 x)^{\frac{1}{x}})

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

Solution steps

\frac{d}{d x} ((3 x)^{\frac{1}{x}})

Apply exponent rule:

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

(3 x)^{\frac{1}{x}} = e^{\frac{1}{x} l n (3 x)}

= \frac{d}{d x} (e^{\frac{1}{x} l n (3 x)})

Apply the chain rule:

e^{\frac{1}{x} l n (3 x)} \frac{d}{d x} (\frac{1}{x} ln (3 x))

= e^{\frac{1}{x} l n (3 x)} \frac{d}{d x} (\frac{1}{x} ln (3 x))

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

= e^{\frac{1}{x} l n (3 x)} \frac{- ln ( 3 x ) + 1}{x ^{2}}

Simplify

e^{\frac{1}{x} l n (3 x)} \frac{- ln ( 3 x ) + 1}{x ^{2}} : 3^{\frac{1}{x}} x^{\frac{1 - 2 x}{x}} (- ln (3 x) + 1)

= 3^{\frac{1}{x}} x^{\frac{1 - 2 x}{x}} (- ln (3 x) + 1)

\frac{d}{d x} (ln (x) tan (x))

3 x^{2} + 4 y (y^{^{'}}) = 0

\frac{d}{d x} (x^{4} - 17 x^{2} + 16)

\int \frac{a ^{x}}{ln ( a )} d x

f (x) = \frac{x}{x + 6}

What is the derivative of (3x^{1/x}) ?
The derivative of (3x^{1/x}) is 3^{1/x}x^{(1-2x)/x}(-ln(3x)+1)
What is the first derivative of (3x^{1/x}) ?
The first derivative of (3x^{1/x}) is 3^{1/x}x^{(1-2x)/x}(-ln(3x)+1)