What is the derivative of log_{2}(((3x+1)/(x^2+1))) ?

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

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

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

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

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

Solution steps

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

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{7 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}})

\int x^{4} e^{- 3 x} d x

\frac{d}{d x} (x (sin (x))^{2})

7

\int_{0}^{1} \frac{x - 9}{x ^{2} - 5 x + 6} d x

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