What is the derivative of (x+4)/(x^3+x-5) ?

The derivative of (x+4)/(x^3+x-5) is (-2x^3-12x^2-9)/((x^3+x-5)^2)

What is the first derivative of (x+4)/(x^3+x-5) ?

The first derivative of (x+4)/(x^3+x-5) is (-2x^3-12x^2-9)/((x^3+x-5)^2)

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derivative of

\frac{x + 4}{x ^{3} + x - 5}

\frac{- 2 x ^{3} - 12 x ^{2} - 9}{( x ^{3} + x - 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x + 4}{x ^{3} + x - 5})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x + 4 ) ( x ^{3} + x - 5 ) - \frac{d}{d x} ( x ^{3} + x - 5 ) ( x + 4 )}{( x ^{3} + x - 5 ) ^{2}}

\frac{d}{d x} (x + 4) = 1

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

= \frac{1 \cdot ( x ^{3} + x - 5 ) - ( 3 x ^{2} + 1 ) ( x + 4 )}{( x ^{3} + x - 5 ) ^{2}}

Simplify

\frac{1 \cdot ( x ^{3} + x - 5 ) - ( 3 x ^{2} + 1 ) ( x + 4 )}{( x ^{3} + x - 5 ) ^{2}} : \frac{- 2 x ^{3} - 12 x ^{2} - 9}{( x ^{3} + x - 5 ) ^{2}}

= \frac{- 2 x ^{3} - 12 x ^{2} - 9}{( x ^{3} + x - 5 ) ^{2}}

y = 2 x, x = 5, y = \frac{2}{x}

\int \frac{x + 1}{x ^{3} + x ^{2} - 6 x} d x

y = x^{2}, at x = - 0.5

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

y = (4 x^{6} + 7)^{5}

What is the derivative of (x+4)/(x^3+x-5) ?
The derivative of (x+4)/(x^3+x-5) is (-2x^3-12x^2-9)/((x^3+x-5)^2)
What is the first derivative of (x+4)/(x^3+x-5) ?
The first derivative of (x+4)/(x^3+x-5) is (-2x^3-12x^2-9)/((x^3+x-5)^2)