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

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

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

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

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = ln (\frac{x}{4}), g = x - 4

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{1}{2} (1 - cos (2 x)))

\int \frac{16}{x ^{2} - 4} d x

- 10 x + 11, x^{2} - 5 x - 13

y = \frac{5 x}{( x + 1 ) ^{2}}, (0, 0)

x y^{^{'}} + y^{2} = 0

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