What is the derivative of y=(x^2+5)/(x-1) ?

The derivative of y=(x^2+5)/(x-1) is (x^2-2x-5)/((x-1)^2)

What is the first derivative of y=(x^2+5)/(x-1) ?

The first derivative of y=(x^2+5)/(x-1) is (x^2-2x-5)/((x-1)^2)

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derivative of

y = \frac{x ^{2} + 5}{x - 1}

\frac{x ^{2} - 2 x - 5}{( x - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 5}{x - 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 5 ) ( x - 1 ) - \frac{d}{d x} ( x - 1 ) ( x ^{2} + 5 )}{( x - 1 ) ^{2}}

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

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

= \frac{2 x ( x - 1 ) - 1 \cdot ( x ^{2} + 5 )}{( x - 1 ) ^{2}}

Simplify

\frac{2 x ( x - 1 ) - 1 \cdot ( x ^{2} + 5 )}{( x - 1 ) ^{2}} : \frac{x ^{2} - 2 x - 5}{( x - 1 ) ^{2}}

= \frac{x ^{2} - 2 x - 5}{( x - 1 ) ^{2}}

\int (x^{2} + 6 x) cos (x) d x

\int \frac{8 + x + x ^{2}}{x} d x

\int_{0}^{\frac{1}{2}} 5 arccos (x) d x

\frac{d}{d x} (e^{a x^{6}})

f (x) = (x)^{3} \cdot e^{(3 x)}

What is the derivative of y=(x^2+5)/(x-1) ?
The derivative of y=(x^2+5)/(x-1) is (x^2-2x-5)/((x-1)^2)
What is the first derivative of y=(x^2+5)/(x-1) ?
The first derivative of y=(x^2+5)/(x-1) is (x^2-2x-5)/((x-1)^2)