What is the derivative of (x^2-6x+1)/(x^2+7) ?

The derivative of (x^2-6x+1)/(x^2+7) is (6x^2+12x-42)/((x^2+7)^2)

What is the first derivative of (x^2-6x+1)/(x^2+7) ?

The first derivative of (x^2-6x+1)/(x^2+7) is (6x^2+12x-42)/((x^2+7)^2)

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derivative of

\frac{x ^{2} - 6 x + 1}{x ^{2} + 7}

\frac{6 x ^{2} + 12 x - 42}{( x ^{2} + 7 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 6 x + 1}{x ^{2} + 7})

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} + 7 ) - \frac{d}{d x} ( x ^{2} + 7 ) ( x ^{2} - 6 x + 1 )}{( x ^{2} + 7 ) ^{2}}

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

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

= \frac{( 2 x - 6 ) ( x ^{2} + 7 ) - 2 x ( x ^{2} - 6 x + 1 )}{( x ^{2} + 7 ) ^{2}}

Expand

(2 x - 6) (x^{2} + 7) - 2 x (x^{2} - 6 x + 1) : 6 x^{2} + 12 x - 42

= \frac{6 x ^{2} + 12 x - 42}{( x ^{2} + 7 ) ^{2}}

\frac{d}{d x} (\frac{1}{2} e^{\frac{x}{2}})

y^{^{''}} + 4 y^{^{'}} = 5

\int \frac{4}{x} + csc (x) cot (x) d x

\frac{d}{d x} (\frac{4 x}{1 + x ^{2}})

\frac{d y}{d x} = \frac{y}{x ^{2}}, y (1) = 1

What is the derivative of (x^2-6x+1)/(x^2+7) ?
The derivative of (x^2-6x+1)/(x^2+7) is (6x^2+12x-42)/((x^2+7)^2)
What is the first derivative of (x^2-6x+1)/(x^2+7) ?
The first derivative of (x^2-6x+1)/(x^2+7) is (6x^2+12x-42)/((x^2+7)^2)