What is the derivative of ln((7x+8/(4x+8))) ?

The derivative of ln((7x+8/(4x+8))) is 6/((7x+8)(x+2))

What is the first derivative of ln((7x+8/(4x+8))) ?

The first derivative of ln((7x+8/(4x+8))) is 6/((7x+8)(x+2))

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\frac{d}{d x} (ln (\frac{7 x + 8}{4 x + 8}))

\frac{6}{( 7 x + 8 ) ( x + 2 )}

Solution steps

\frac{d}{d x} (ln (\frac{7 x + 8}{4 x + 8}))

Apply the chain rule:

\frac{1}{\frac{7 x + 8}{4 x + 8}} \frac{d}{d x} (\frac{7 x + 8}{4 x + 8})

= \frac{1}{\frac{7 x + 8}{4 x + 8}} \frac{d}{d x} (\frac{7 x + 8}{4 x + 8})

\frac{d}{d x} (\frac{7 x + 8}{4 x + 8}) = \frac{3}{2 ( x + 2 ) ^{2}}

= \frac{1}{\frac{7 x + 8}{4 x + 8}} \cdot \frac{3}{2 ( x + 2 ) ^{2}}

Simplify

\frac{1}{\frac{7 x + 8}{4 x + 8}} \cdot \frac{3}{2 ( x + 2 ) ^{2}} : \frac{6}{( 7 x + 8 ) ( x + 2 )}

= \frac{6}{( 7 x + 8 ) ( x + 2 )}

y = \frac{( 3 x ^{2} - x - 2 )}{3 x ^{2}}

\int \frac{cos ^{3} ( x )}{sin ^{4} ( x )} d x

\int_{- 6}^{4} ∣ x ∣ d x

t \to 1 lim (\frac{t - 1}{t ^{2} - 1})

\frac{x ^{3}}{3}

What is the derivative of ln((7x+8/(4x+8))) ?
The derivative of ln((7x+8/(4x+8))) is 6/((7x+8)(x+2))
What is the first derivative of ln((7x+8/(4x+8))) ?
The first derivative of ln((7x+8/(4x+8))) is 6/((7x+8)(x+2))