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

The derivative of y=(x+3)/(x^3+x-1) is (-2x^3-9x^2-4)/((x^3+x-1)^2)

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

The first derivative of y=(x+3)/(x^3+x-1) is (-2x^3-9x^2-4)/((x^3+x-1)^2)

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derivative of

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

\frac{- 2 x ^{3} - 9 x ^{2} - 4}{( x ^{3} + x - 1 ) ^{2}}

Solution steps

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

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

Simplify

\frac{1 \cdot ( x ^{3} + x - 1 ) - ( 3 x ^{2} + 1 ) ( x + 3 )}{( x ^{3} + x - 1 ) ^{2}} : \frac{- 2 x ^{3} - 9 x ^{2} - 4}{( x ^{3} + x - 1 ) ^{2}}

= \frac{- 2 x ^{3} - 9 x ^{2} - 4}{( x ^{3} + x - 1 ) ^{2}}

\frac{\partial}{\partial x} (14 - x^{2} - 4 x y)

\int x^{2} 8 x - 7 d x

\int_{1}^{5} \frac{x}{e ^{x}} d x

x^{2} - 6 x

f (x) = \frac{3}{x - 2}

What is the derivative of y=(x+3)/(x^3+x-1) ?
The derivative of y=(x+3)/(x^3+x-1) is (-2x^3-9x^2-4)/((x^3+x-1)^2)
What is the first derivative of y=(x+3)/(x^3+x-1) ?
The first derivative of y=(x+3)/(x^3+x-1) is (-2x^3-9x^2-4)/((x^3+x-1)^2)