What is the derivative of ln(x/(4x^2+x)) ?

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

What is the first derivative of ln(x/(4x^2+x)) ?

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

Popular >

derivative of

ln (\frac{x}{4 x ^{2} + x})

- \frac{4}{4 x + 1}

Solution steps

\frac{d}{d x} (ln (\frac{x}{4 x ^{2} + x}))

Simplify

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

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

Take the constant out:

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

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

Apply the chain rule:

\frac{1}{4 x + 1} \frac{d}{d x} (4 x + 1)

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

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

= - \frac{1}{4 x + 1} \cdot 4

Simplify

- \frac{1}{4 x + 1} \cdot 4 : - \frac{4}{4 x + 1}

= - \frac{4}{4 x + 1}

\frac{d}{d x} (\frac{1}{2} \cdot 3 x)

\int_{0}^{1} e^{- 3 x} d x

\frac{\partial}{\partial x} (\frac{1}{y} - 2 x sin (y))

\frac{d}{d x} (\frac{( e ^{x} - 1 )}{x})

y^{^{''}} - y^{^{'}} - 2 y = - 2 t + 6 t^{2}

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