What is the derivative of (4x/((x-1)^2)) ?

The derivative of (4x/((x-1)^2)) is (4(-1-x))/((x-1)^3)

What is the first derivative of (4x/((x-1)^2)) ?

The first derivative of (4x/((x-1)^2)) is (4(-1-x))/((x-1)^3)

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

\frac{4 ( - 1 - x )}{( x - 1 ) ^{3}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

= 4 \cdot \frac{\frac{d x}{d x} ( x - 1 ) ^{2} - \frac{d}{d x} ( ( x - 1 ) ^{2} ) x}{( ( x - 1 ) ^{2} ) ^{2}}

\frac{d x}{d x} = 1

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

= 4 \cdot \frac{1 \cdot ( x - 1 ) ^{2} - 2 ( x - 1 ) x}{( ( x - 1 ) ^{2} ) ^{2}}

Simplify

4 \cdot \frac{1 \cdot ( x - 1 ) ^{2} - 2 ( x - 1 ) x}{( ( x - 1 ) ^{2} ) ^{2}} : \frac{4 ( - 1 - x )}{( x - 1 ) ^{3}}

= \frac{4 ( - 1 - x )}{( x - 1 ) ^{3}}

\int_{0}^{2} 2 x^{2} + 1 d x

\int_{0}^{1} \frac{x}{1 + x ^{2}} d x

\int \frac{( ln ( x ) ) ^{36}}{x} d x

\int_{- 1}^{1} (6 - 6 x^{2})^{2} d x

What is the derivative of (4x/((x-1)^2)) ?
The derivative of (4x/((x-1)^2)) is (4(-1-x))/((x-1)^3)
What is the first derivative of (4x/((x-1)^2)) ?
The first derivative of (4x/((x-1)^2)) is (4(-1-x))/((x-1)^3)