What is the derivative of (x^4-x^3-8/((x-3)(x+2))) ?

The derivative of (x^4-x^3-8/((x-3)(x+2))) is (2x^5-4x^4-22x^3+18x^2+16x-8)/((x-3)^2(x+2)^2)

What is the first derivative of (x^4-x^3-8/((x-3)(x+2))) ?

The first derivative of (x^4-x^3-8/((x-3)(x+2))) is (2x^5-4x^4-22x^3+18x^2+16x-8)/((x-3)^2(x+2)^2)

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\frac{d}{d x} (\frac{x ^{4} - x ^{3} - 8}{( x - 3 ) ( x + 2 )})

\frac{2 x ^{5} - 4 x ^{4} - 22 x ^{3} + 18 x ^{2} + 16 x - 8}{( x - 3 ) ^{2} ( x + 2 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{4} - x ^{3} - 8}{( 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 ^{4} - x ^{3} - 8 ) ( x - 3 ) ( x + 2 ) - \frac{d}{d x} ( ( x - 3 ) ( x + 2 ) ) ( x ^{4} - x ^{3} - 8 )}{( ( x - 3 ) ( x + 2 ) ) ^{2}}

\frac{d}{d x} (x^{4} - x^{3} - 8) = 4 x^{3} - 3 x^{2}

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

= \frac{( 4 x ^{3} - 3 x ^{2} ) ( x - 3 ) ( x + 2 ) - ( 2 x - 1 ) ( x ^{4} - x ^{3} - 8 )}{( ( x - 3 ) ( x + 2 ) ) ^{2}}

Simplify

\frac{( 4 x ^{3} - 3 x ^{2} ) ( x - 3 ) ( x + 2 ) - ( 2 x - 1 ) ( x ^{4} - x ^{3} - 8 )}{( ( x - 3 ) ( x + 2 ) ) ^{2}} : \frac{2 x ^{5} - 4 x ^{4} - 22 x ^{3} + 18 x ^{2} + 16 x - 8}{( x - 3 ) ^{2} ( x + 2 ) ^{2}}

= \frac{2 x ^{5} - 4 x ^{4} - 22 x ^{3} + 18 x ^{2} + 16 x - 8}{( x - 3 ) ^{2} ( x + 2 ) ^{2}}

y = x^{3} - 4, (2, 0)

\int_{0}^{2} (3 - x^{3}) d x

\int \frac{36 + ( ln ( 4 x ) ) ^{2}}{x} d x

\int \frac{5 t t}{1 + t ^{5}} d t

y = sin (x) \cdot cos (x)

What is the derivative of (x^4-x^3-8/((x-3)(x+2))) ?
The derivative of (x^4-x^3-8/((x-3)(x+2))) is (2x^5-4x^4-22x^3+18x^2+16x-8)/((x-3)^2(x+2)^2)
What is the first derivative of (x^4-x^3-8/((x-3)(x+2))) ?
The first derivative of (x^4-x^3-8/((x-3)(x+2))) is (2x^5-4x^4-22x^3+18x^2+16x-8)/((x-3)^2(x+2)^2)