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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

\frac{d}{d x} (x sin (x + y))

\int \frac{x}{x ^{2} + 2 x + 25} d x

\int \frac{5 x + 7}{4 x ^{2} + 12 x + 10} d x

f (t) = \frac{( e ^{x} + 5 x )}{5 x}

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