What is the derivative of (x^2-9x+14/((x-5)(x-10))) ?

The derivative of (x^2-9x+14/((x-5)(x-10))) is (-6x^2+72x-240)/((x-5)^2(x-10)^2)

What is the first derivative of (x^2-9x+14/((x-5)(x-10))) ?

The first derivative of (x^2-9x+14/((x-5)(x-10))) is (-6x^2+72x-240)/((x-5)^2(x-10)^2)

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\frac{d}{d x} (\frac{x ^{2} - 9 x + 14}{( x - 5 ) ( x - 10 )})

\frac{- 6 x ^{2} + 72 x - 240}{( x - 5 ) ^{2} ( x - 10 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 9 x + 14}{( x - 5 ) ( x - 10 )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 9 x + 14 ) ( x - 5 ) ( x - 10 ) - \frac{d}{d x} ( ( x - 5 ) ( x - 10 ) ) ( x ^{2} - 9 x + 14 )}{( ( x - 5 ) ( x - 10 ) ) ^{2}}

\frac{d}{d x} (x^{2} - 9 x + 14) = 2 x - 9

\frac{d}{d x} ((x - 5) (x - 10)) = 2 x - 15

= \frac{( 2 x - 9 ) ( x - 5 ) ( x - 10 ) - ( 2 x - 15 ) ( x ^{2} - 9 x + 14 )}{( ( x - 5 ) ( x - 10 ) ) ^{2}}

Simplify

\frac{( 2 x - 9 ) ( x - 5 ) ( x - 10 ) - ( 2 x - 15 ) ( x ^{2} - 9 x + 14 )}{( ( x - 5 ) ( x - 10 ) ) ^{2}} : \frac{- 6 x ^{2} + 72 x - 240}{( x - 5 ) ^{2} ( x - 10 ) ^{2}}

= \frac{- 6 x ^{2} + 72 x - 240}{( x - 5 ) ^{2} ( x - 10 ) ^{2}}

\frac{\partial}{\partial x} (y x^{- 1})

\frac{( x ^{2} + 1 ) d y}{d x} + 6 x (y - 1) = 0, y (0) = 6

\int \frac{1}{x ^{3}} ln (x) d x

\frac{\partial}{\partial y} (x^{2})

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

What is the derivative of (x^2-9x+14/((x-5)(x-10))) ?
The derivative of (x^2-9x+14/((x-5)(x-10))) is (-6x^2+72x-240)/((x-5)^2(x-10)^2)
What is the first derivative of (x^2-9x+14/((x-5)(x-10))) ?
The first derivative of (x^2-9x+14/((x-5)(x-10))) is (-6x^2+72x-240)/((x-5)^2(x-10)^2)