What is the derivative of (ln(3x+1)/(6x+2)) ?

The derivative of (ln(3x+1)/(6x+2)) is -(3(ln(3x+1)-1))/(2(3x+1)^2)

What is the first derivative of (ln(3x+1)/(6x+2)) ?

The first derivative of (ln(3x+1)/(6x+2)) is -(3(ln(3x+1)-1))/(2(3x+1)^2)

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\frac{d}{d x} (\frac{ln ( 3 x + 1 )}{6 x + 2})

- \frac{3 ( ln ( 3 x + 1 ) - 1 )}{2 ( 3 x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{ln ( 3 x + 1 )}{6 x + 2})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( ln ( 3 x + 1 ) ) ( 6 x + 2 ) - \frac{d}{d x} ( 6 x + 2 ) ln ( 3 x + 1 )}{( 6 x + 2 ) ^{2}}

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

\frac{d}{d x} (6 x + 2) = 6

= \frac{\frac{3}{3 x + 1} ( 6 x + 2 ) - 6 ln ( 3 x + 1 )}{( 6 x + 2 ) ^{2}}

Simplify

\frac{\frac{3}{3 x + 1} ( 6 x + 2 ) - 6 ln ( 3 x + 1 )}{( 6 x + 2 ) ^{2}} : - \frac{3 ( ln ( 3 x + 1 ) - 1 )}{2 ( 3 x + 1 ) ^{2}}

= - \frac{3 ( ln ( 3 x + 1 ) - 1 )}{2 ( 3 x + 1 ) ^{2}}

\frac{9}{sin ( x ) + cos ( x )}

x, 3 x, 4 - x

\int 4 - y^{2} d y

\int_{0}^{1} \frac{1}{x ( 1 - x )} d x

(7.4) (9.8)

What is the derivative of (ln(3x+1)/(6x+2)) ?
The derivative of (ln(3x+1)/(6x+2)) is -(3(ln(3x+1)-1))/(2(3x+1)^2)
What is the first derivative of (ln(3x+1)/(6x+2)) ?
The first derivative of (ln(3x+1)/(6x+2)) is -(3(ln(3x+1)-1))/(2(3x+1)^2)