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

The derivative of ln((x^3+1/(x^2+1))) is (x^4+3x^2-2x)/((x^3+1)(x^2+1))

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

The first derivative of ln((x^3+1/(x^2+1))) is (x^4+3x^2-2x)/((x^3+1)(x^2+1))

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

\frac{x ^{4} + 3 x ^{2} - 2 x}{( x ^{3} + 1 ) ( x ^{2} + 1 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{3} + 1}{x ^{2} + 1}) = \frac{x ^{4} + 3 x ^{2} - 2 x}{( x ^{2} + 1 ) ^{2}}

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

Simplify

\frac{1}{\frac{x ^{3} + 1}{x ^{2} + 1}} \cdot \frac{x ^{4} + 3 x ^{2} - 2 x}{( x ^{2} + 1 ) ^{2}} : \frac{x ^{4} + 3 x ^{2} - 2 x}{( x ^{3} + 1 ) ( x ^{2} + 1 )}

= \frac{x ^{4} + 3 x ^{2} - 2 x}{( x ^{3} + 1 ) ( x ^{2} + 1 )}

\int 3 e^{3 t} d t

\frac{x}{x + 1}

\int \frac{20 x ^{3} - 40 x ^{2} + 3}{x ^{2} - 2 x} d x

f (x) = \frac{x}{x - 6}

f (x) = 5 x^{2} + 7 x

What is the derivative of ln((x^3+1/(x^2+1))) ?
The derivative of ln((x^3+1/(x^2+1))) is (x^4+3x^2-2x)/((x^3+1)(x^2+1))
What is the first derivative of ln((x^3+1/(x^2+1))) ?
The first derivative of ln((x^3+1/(x^2+1))) is (x^4+3x^2-2x)/((x^3+1)(x^2+1))