What is the derivative of (x^2-6x+1)/(x^2+8) ?

The derivative of (x^2-6x+1)/(x^2+8) is (6x^2+14x-48)/((x^2+8)^2)

What is the first derivative of (x^2-6x+1)/(x^2+8) ?

The first derivative of (x^2-6x+1)/(x^2+8) is (6x^2+14x-48)/((x^2+8)^2)

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derivative of

\frac{x ^{2} - 6 x + 1}{x ^{2} + 8}

\frac{6 x ^{2} + 14 x - 48}{( x ^{2} + 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 6 x + 1}{x ^{2} + 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 6 x + 1 ) ( x ^{2} + 8 ) - \frac{d}{d x} ( x ^{2} + 8 ) ( x ^{2} - 6 x + 1 )}{( x ^{2} + 8 ) ^{2}}

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

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

= \frac{( 2 x - 6 ) ( x ^{2} + 8 ) - 2 x ( x ^{2} - 6 x + 1 )}{( x ^{2} + 8 ) ^{2}}

Expand

(2 x - 6) (x^{2} + 8) - 2 x (x^{2} - 6 x + 1) : 6 x^{2} + 14 x - 48

= \frac{6 x ^{2} + 14 x - 48}{( x ^{2} + 8 ) ^{2}}

5 + 7 x

ln (x e^{x} + 9)

f (x) = \frac{1}{x ^{2}}, at x = - 1

x \to \infty lim (ln (5))

\int ln (6 x^{4}) d x

What is the derivative of (x^2-6x+1)/(x^2+8) ?
The derivative of (x^2-6x+1)/(x^2+8) is (6x^2+14x-48)/((x^2+8)^2)
What is the first derivative of (x^2-6x+1)/(x^2+8) ?
The first derivative of (x^2-6x+1)/(x^2+8) is (6x^2+14x-48)/((x^2+8)^2)