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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

y = 4 - x^{2}, y = x + 2

\frac{\partial}{\partial y} (\frac{1}{x y})

\int \frac{x}{1 + x ^{2}} d x

\int \frac{3}{x ^{2}} - \frac{2}{x ^{3}} d x

x^{3}

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