What is the derivative of y=(x^2+7x-2)/(x-2) ?

The derivative of y=(x^2+7x-2)/(x-2) is (x^2-4x-12)/((x-2)^2)

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

The first derivative of y=(x^2+7x-2)/(x-2) is (x^2-4x-12)/((x-2)^2)

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derivative of

y = \frac{x ^{2} + 7 x - 2}{x - 2}

\frac{x ^{2} - 4 x - 12}{( x - 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 2 x + 7 ) ( x - 2 ) - 1 \cdot ( x ^{2} + 7 x - 2 )}{( x - 2 ) ^{2}} : \frac{x ^{2} - 4 x - 12}{( x - 2 ) ^{2}}

= \frac{x ^{2} - 4 x - 12}{( x - 2 ) ^{2}}

\frac{d x}{d t} = 4 (x^{2} + 1), x (\frac{π}{4}) = 1

\int tan^{4} (2 x) d x

t^{5} y^{4} - t^{3} y^{^{'}} + 6 y = 0

y^{^{'}} = \frac{3 ( e ^{x} + 1 )}{2 y}

\frac{d}{dd} (\frac{a x + b}{c x + d})

What is the derivative of y=(x^2+7x-2)/(x-2) ?
The derivative of y=(x^2+7x-2)/(x-2) is (x^2-4x-12)/((x-2)^2)
What is the first derivative of y=(x^2+7x-2)/(x-2) ?
The first derivative of y=(x^2+7x-2)/(x-2) is (x^2-4x-12)/((x-2)^2)