What is the derivative of (x^2-6x+1)/(x^2+9) ?

The derivative of (x^2-6x+1)/(x^2+9) is (6x^2+16x-54)/((x^2+9)^2)

What is the first derivative of (x^2-6x+1)/(x^2+9) ?

The first derivative of (x^2-6x+1)/(x^2+9) is (6x^2+16x-54)/((x^2+9)^2)

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derivative of

\frac{x ^{2} - 6 x + 1}{x ^{2} + 9}

\frac{6 x ^{2} + 16 x - 54}{( x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 6 x + 1}{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} - 6 x + 1 ) ( x ^{2} + 9 ) - \frac{d}{d x} ( x ^{2} + 9 ) ( x ^{2} - 6 x + 1 )}{( x ^{2} + 9 ) ^{2}}

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

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

= \frac{( 2 x - 6 ) ( x ^{2} + 9 ) - 2 x ( x ^{2} - 6 x + 1 )}{( x ^{2} + 9 ) ^{2}}

Expand

(2 x - 6) (x^{2} + 9) - 2 x (x^{2} - 6 x + 1) : 6 x^{2} + 16 x - 54

= \frac{6 x ^{2} + 16 x - 54}{( x ^{2} + 9 ) ^{2}}

12 x^{11} y + x^{12} y^{^{'}} = - 7 sec^{2} (x) \cdot tan (x)

\frac{d}{d x} ((2 x - e^{2} x)^{s i n (2 x)})

\int \frac{x}{t ^{2}} d t

4 x - 8

What is the derivative of (x^2-6x+1)/(x^2+9) ?
The derivative of (x^2-6x+1)/(x^2+9) is (6x^2+16x-54)/((x^2+9)^2)
What is the first derivative of (x^2-6x+1)/(x^2+9) ?
The first derivative of (x^2-6x+1)/(x^2+9) is (6x^2+16x-54)/((x^2+9)^2)