What is the derivative of f(x)=(x^2)/(x-1) ?

The derivative of f(x)=(x^2)/(x-1) is (x^2-2x)/((x-1)^2)

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

The first derivative of f(x)=(x^2)/(x-1) is (x^2-2x)/((x-1)^2)

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

4 (8 x - 1)^{4}

x^{^{''}} + 4 x^{^{'}} + 4 x = e^{- 2 t}

f (x) = 3 x^{3}

\int_{0}^{8} - x^{2} + 2 x + 4 d x

f (x) = \frac{4 x}{x + 1}, at x = 3

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