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

The derivative of ((x^3-3))/((x^2+x-1)) is (x^4+2x^3-3x^2+6x+3)/((x^2+x-1)^2)

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

The first derivative of ((x^3-3))/((x^2+x-1)) is (x^4+2x^3-3x^2+6x+3)/((x^2+x-1)^2)

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derivative of

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

\frac{x ^{4} + 2 x ^{3} - 3 x ^{2} + 6 x + 3}{( x ^{2} + x - 1 ) ^{2}}

Solution steps

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

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

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

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

Expand

3 x^{2} (x^{2} + x - 1) - (2 x + 1) (x^{3} - 3) : x^{4} + 2 x^{3} - 3 x^{2} + 6 x + 3

= \frac{x ^{4} + 2 x ^{3} - 3 x ^{2} + 6 x + 3}{( x ^{2} + x - 1 ) ^{2}}

(2 + x^{2}) y^{^{'}} - 2 x y = 0

y = 8 x - x^{2}

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

\frac{d}{d x} (\frac{( x - 2 )}{x + 2})

y^{^{''}} + 30 y = 0

What is the derivative of ((x^3-3))/((x^2+x-1)) ?
The derivative of ((x^3-3))/((x^2+x-1)) is (x^4+2x^3-3x^2+6x+3)/((x^2+x-1)^2)
What is the first derivative of ((x^3-3))/((x^2+x-1)) ?
The first derivative of ((x^3-3))/((x^2+x-1)) is (x^4+2x^3-3x^2+6x+3)/((x^2+x-1)^2)