What is the derivative of-4ln(+1/x-4) ?

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(x^{2} + 2 y^{2}) d x = x y d y

\int \frac{sin ( \frac{1}{x ^{4}} )}{x ^{5}} d x

\int_{0.2}^{0.4} 150 x d x

\frac{5}{x + 5}

y^{^{'}} - (2 x + 1) y = 0

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