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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (e^{4 x y})

\int cos^{3} (5 x) sin^{- 2} (5 x) d x

\int 2 e^{- 3 x} d x

x \to 0 lim (cos^{\frac{1}{x}} (x))

sin^{2} (y) d x + cos^{2} (x) d y = 0

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