What is the derivative of 1/8 ln((3x-4/(3x+4))) ?

The derivative of 1/8 ln((3x-4/(3x+4))) is 3/((3x-4)(3x+4))

What is the first derivative of 1/8 ln((3x-4/(3x+4))) ?

The first derivative of 1/8 ln((3x-4/(3x+4))) is 3/((3x-4)(3x+4))

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\frac{d}{d x} (\frac{1}{8} ln (\frac{3 x - 4}{3 x + 4}))

\frac{3}{( 3 x - 4 ) ( 3 x + 4 )}

Solution steps

\frac{d}{d x} (\frac{1}{8} ln (\frac{3 x - 4}{3 x + 4}))

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (ln (\frac{3 x - 4}{3 x + 4}))

Apply the chain rule:

\frac{1}{\frac{3 x - 4}{3 x + 4}} \frac{d}{d x} (\frac{3 x - 4}{3 x + 4})

= \frac{1}{\frac{3 x - 4}{3 x + 4}} \frac{d}{d x} (\frac{3 x - 4}{3 x + 4})

\frac{d}{d x} (\frac{3 x - 4}{3 x + 4}) = \frac{24}{( 3 x + 4 ) ^{2}}

= \frac{1}{8} \cdot \frac{1}{\frac{3 x - 4}{3 x + 4}} \cdot \frac{24}{( 3 x + 4 ) ^{2}}

Simplify

\frac{1}{8} \cdot \frac{1}{\frac{3 x - 4}{3 x + 4}} \cdot \frac{24}{( 3 x + 4 ) ^{2}} : \frac{3}{( 3 x - 4 ) ( 3 x + 4 )}

= \frac{3}{( 3 x - 4 ) ( 3 x + 4 )}

\frac{\partial}{\partial x} (- x^{3} + 4 x y - 2 y^{2} + 1)

x \to 7 lim (2 x + 5)

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

\int (5 x - 1)^{7} d x

2 x, 9 x - x^{2}, (7, 14)

What is the derivative of 1/8 ln((3x-4/(3x+4))) ?
The derivative of 1/8 ln((3x-4/(3x+4))) is 3/((3x-4)(3x+4))
What is the first derivative of 1/8 ln((3x-4/(3x+4))) ?
The first derivative of 1/8 ln((3x-4/(3x+4))) is 3/((3x-4)(3x+4))