What is the derivative of (x^2+3/(x^2-9x)) ?

The derivative of (x^2+3/(x^2-9x)) is (-9x^2-6x+27)/((x^2-9x)^2)

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

The first derivative of (x^2+3/(x^2-9x)) is (-9x^2-6x+27)/((x^2-9x)^2)

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

\frac{- 9 x ^{2} - 6 x + 27}{( x ^{2} - 9 x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

2 x (x^{2} - 9 x) - (2 x - 9) (x^{2} + 3) : - 9 x^{2} - 6 x + 27

= \frac{- 9 x ^{2} - 6 x + 27}{( x ^{2} - 9 x ) ^{2}}

\frac{d}{d x} (\frac{ln ( 2 x )}{2 x})

\frac{\partial}{\partial u} (u \cdot sin (v))

\int_{- 2}^{1} \frac{1}{2 - x} d x

\frac{d}{d x} (x e^{- 6 x})

tan (e^{5 t}) + e^{t a n (5 t)}

What is the derivative of (x^2+3/(x^2-9x)) ?
The derivative of (x^2+3/(x^2-9x)) is (-9x^2-6x+27)/((x^2-9x)^2)
What is the first derivative of (x^2+3/(x^2-9x)) ?
The first derivative of (x^2+3/(x^2-9x)) is (-9x^2-6x+27)/((x^2-9x)^2)