What is the derivative of (x^2-9x+7/(x^2+9)) ?

The derivative of (x^2-9x+7/(x^2+9)) is (9x^2+4x-81)/((x^2+9)^2)

What is the first derivative of (x^2-9x+7/(x^2+9)) ?

The first derivative of (x^2-9x+7/(x^2+9)) is (9x^2+4x-81)/((x^2+9)^2)

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\frac{d}{d x} (\frac{x ^{2} - 9 x + 7}{x ^{2} + 9})

\frac{9 x ^{2} + 4 x - 81}{( x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 9 x + 7}{x ^{2} + 9})

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 + 7 ) ( x ^{2} + 9 ) - \frac{d}{d x} ( x ^{2} + 9 ) ( x ^{2} - 9 x + 7 )}{( x ^{2} + 9 ) ^{2}}

\frac{d}{d x} (x^{2} - 9 x + 7) = 2 x - 9

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

= \frac{( 2 x - 9 ) ( x ^{2} + 9 ) - 2 x ( x ^{2} - 9 x + 7 )}{( x ^{2} + 9 ) ^{2}}

Expand

(2 x - 9) (x^{2} + 9) - 2 x (x^{2} - 9 x + 7) : 9 x^{2} + 4 x - 81

= \frac{9 x ^{2} + 4 x - 81}{( x ^{2} + 9 ) ^{2}}

10 tan^{2} (6 x)

\frac{d}{d x} (2 sin (2 x) - 4 sin^{9} (2 x))

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

19 - x^{2}, - 2 x + 16, - 5, - 19

\frac{\partial}{\partial x} (y \cdot ∣ x ∣)

What is the derivative of (x^2-9x+7/(x^2+9)) ?
The derivative of (x^2-9x+7/(x^2+9)) is (9x^2+4x-81)/((x^2+9)^2)
What is the first derivative of (x^2-9x+7/(x^2+9)) ?
The first derivative of (x^2-9x+7/(x^2+9)) is (9x^2+4x-81)/((x^2+9)^2)