What is the derivative of (x^2-3x+8)/(x^2+9) ?

The derivative of (x^2-3x+8)/(x^2+9) is (3x^2+2x-27)/((x^2+9)^2)

What is the first derivative of (x^2-3x+8)/(x^2+9) ?

The first derivative of (x^2-3x+8)/(x^2+9) is (3x^2+2x-27)/((x^2+9)^2)

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derivative of

\frac{x ^{2} - 3 x + 8}{x ^{2} + 9}

\frac{3 x ^{2} + 2 x - 27}{( x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 3 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} - 3 x + 8 ) ( x ^{2} + 9 ) - \frac{d}{d x} ( x ^{2} + 9 ) ( x ^{2} - 3 x + 8 )}{( x ^{2} + 9 ) ^{2}}

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

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

= \frac{( 2 x - 3 ) ( x ^{2} + 9 ) - 2 x ( x ^{2} - 3 x + 8 )}{( x ^{2} + 9 ) ^{2}}

Expand

(2 x - 3) (x^{2} + 9) - 2 x (x^{2} - 3 x + 8) : 3 x^{2} + 2 x - 27

= \frac{3 x ^{2} + 2 x - 27}{( x ^{2} + 9 ) ^{2}}

x^{4 x}

\frac{1}{s - 1} + \frac{1}{s + 3}

f (x) = x^{3} + x^{2} - x - 3, at x = - 2

3 x^{2} - 4 x + 5

f (x) = 3 sin (x) - 33 cos (x)

What is the derivative of (x^2-3x+8)/(x^2+9) ?
The derivative of (x^2-3x+8)/(x^2+9) is (3x^2+2x-27)/((x^2+9)^2)
What is the first derivative of (x^2-3x+8)/(x^2+9) ?
The first derivative of (x^2-3x+8)/(x^2+9) is (3x^2+2x-27)/((x^2+9)^2)