What is the derivative of (x^2-6x+5/(x^2+7)) ?

The derivative of (x^2-6x+5/(x^2+7)) is (6x^2+4x-42)/((x^2+7)^2)

What is the first derivative of (x^2-6x+5/(x^2+7)) ?

The first derivative of (x^2-6x+5/(x^2+7)) is (6x^2+4x-42)/((x^2+7)^2)

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\frac{d}{d x} (\frac{x ^{2} - 6 x + 5}{x ^{2} + 7})

\frac{6 x ^{2} + 4 x - 42}{( x ^{2} + 7 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 6 x + 5}{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 + 5 ) ( x ^{2} + 7 ) - \frac{d}{d x} ( x ^{2} + 7 ) ( x ^{2} - 6 x + 5 )}{( x ^{2} + 7 ) ^{2}}

\frac{d}{d x} (x^{2} - 6 x + 5) = 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 + 5 )}{( x ^{2} + 7 ) ^{2}}

Expand

(2 x - 6) (x^{2} + 7) - 2 x (x^{2} - 6 x + 5) : 6 x^{2} + 4 x - 42

= \frac{6 x ^{2} + 4 x - 42}{( x ^{2} + 7 ) ^{2}}

\int_{0}^{3} (3 x^{2} + x - 2) d x

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{- 1} e^{2 t}

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

\frac{\partial}{\partial x} (x^{2} (x - y)^{- 2})

What is the derivative of (x^2-6x+5/(x^2+7)) ?
The derivative of (x^2-6x+5/(x^2+7)) is (6x^2+4x-42)/((x^2+7)^2)
What is the first derivative of (x^2-6x+5/(x^2+7)) ?
The first derivative of (x^2-6x+5/(x^2+7)) is (6x^2+4x-42)/((x^2+7)^2)