What is the derivative of (ln(x)/(4x)) ?

The derivative of (ln(x)/(4x)) is (1-ln(x))/(4x^2)

What is the first derivative of (ln(x)/(4x)) ?

The first derivative of (ln(x)/(4x)) is (1-ln(x))/(4x^2)

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\frac{d}{d x} (\frac{ln ( x )}{4 x})

\frac{1 - ln ( x )}{4 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{ln ( x )}{4 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{4} \frac{d}{d x} (\frac{ln ( x )}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{4} \cdot \frac{\frac{d}{d x} ( ln ( x ) ) x - \frac{d x}{d x} ln ( x )}{x ^{2}}

\frac{d}{d x} (ln (x)) = \frac{1}{x}

\frac{d x}{d x} = 1

= \frac{1}{4} \cdot \frac{\frac{1}{x} x - 1 \cdot ln ( x )}{x ^{2}}

Simplify

\frac{1}{4} \cdot \frac{\frac{1}{x} x - 1 \cdot ln ( x )}{x ^{2}} : \frac{1 - ln ( x )}{4 x ^{2}}

= \frac{1 - ln ( x )}{4 x ^{2}}

\frac{6}{x + 4}

x \to - 2 - lim (ln (x + 3))

\frac{2}{\frac{1}{4} + h} - 4

\int \frac{x}{( x ^{2} + 2 x + 2 )} d x

y = x^{4} e^{- x}, (1, \frac{1}{e})

What is the derivative of (ln(x)/(4x)) ?
The derivative of (ln(x)/(4x)) is (1-ln(x))/(4x^2)
What is the first derivative of (ln(x)/(4x)) ?
The first derivative of (ln(x)/(4x)) is (1-ln(x))/(4x^2)