What is the derivative of (x^2-9x+8)/(x^2+9) ?

The derivative of (x^2-9x+8)/(x^2+9) is (9x^2+2x-81)/((x^2+9)^2)

What is the first derivative of (x^2-9x+8)/(x^2+9) ?

The first derivative of (x^2-9x+8)/(x^2+9) is (9x^2+2x-81)/((x^2+9)^2)

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derivative of

\frac{x ^{2} - 9 x + 8}{x ^{2} + 9}

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

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 9 x + 8}{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 + 8 ) ( x ^{2} + 9 ) - \frac{d}{d x} ( x ^{2} + 9 ) ( x ^{2} - 9 x + 8 )}{( x ^{2} + 9 ) ^{2}}

\frac{d}{d x} (x^{2} - 9 x + 8) = 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 + 8 )}{( x ^{2} + 9 ) ^{2}}

Expand

(2 x - 9) (x^{2} + 9) - 2 x (x^{2} - 9 x + 8) : 9 x^{2} + 2 x - 81

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

n = 1 \sum \infty \frac{8}{n} - \frac{8}{n + 1}

\int \frac{6 x - 11}{( x - 1 ) ^{2}} d x

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

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

\frac{6}{( s + 3 ) ^{2}}

What is the derivative of (x^2-9x+8)/(x^2+9) ?
The derivative of (x^2-9x+8)/(x^2+9) is (9x^2+2x-81)/((x^2+9)^2)
What is the first derivative of (x^2-9x+8)/(x^2+9) ?
The first derivative of (x^2-9x+8)/(x^2+9) is (9x^2+2x-81)/((x^2+9)^2)