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

The derivative of ln(((x+2e^{3x})/(x^2+1))) is (3x^3+5x^2-x+7)/((x+2)(x^2+1))

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

The first derivative of ln(((x+2e^{3x})/(x^2+1))) is (3x^3+5x^2-x+7)/((x+2)(x^2+1))

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

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

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{\frac{( x + 2 ) e ^{3 x}}{x ^{2} + 1}} \cdot \frac{3 e ^{3 x} x ^{3} + 5 e ^{3 x} x ^{2} - e ^{3 x} x + 7 e ^{3 x}}{( x ^{2} + 1 ) ^{2}}

Simplify

\frac{1}{\frac{( x + 2 ) e ^{3 x}}{x ^{2} + 1}} \cdot \frac{3 e ^{3 x} x ^{3} + 5 e ^{3 x} x ^{2} - e ^{3 x} x + 7 e ^{3 x}}{( x ^{2} + 1 ) ^{2}} : \frac{3 x ^{3} + 5 x ^{2} - x + 7}{( x + 2 ) ( x ^{2} + 1 )}

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

\frac{d y}{d x} = (\frac{9 x}{9 x ^{2} - 1}) e^{- y}

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

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

x^{4} - 3 x^{2} + 1, 1

y = e^{x} (x + x x)

What is the derivative of ln(((x+2e^{3x})/(x^2+1))) ?
The derivative of ln(((x+2e^{3x})/(x^2+1))) is (3x^3+5x^2-x+7)/((x+2)(x^2+1))
What is the first derivative of ln(((x+2e^{3x})/(x^2+1))) ?
The first derivative of ln(((x+2e^{3x})/(x^2+1))) is (3x^3+5x^2-x+7)/((x+2)(x^2+1))