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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{2 x ^{2} ( x ^{2} - 2 x )}{( x - 1 ) ^{3}}

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

\int e^{- 4}

f (x) = - x^{3} + 4 x^{2} - 1, at x = 1

\frac{d}{d x} (e^{- 3 x + 4})

f (x) = x^{23} + 50 x^{17} + 223

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