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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

2 (t - \frac{π}{4}) sin (4 t)

3 x^{2} + 3 x

\frac{d}{d t} (\frac{1}{t ^{2} + 1})

f (x) = x + 1 - ln (x), at x = 0.5

(t^{2} + x^{2}) d t + (t x) d x = 0

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