What is the derivative of ((-3x^2)/(((x^2-2x+1)))) ?

The derivative of ((-3x^2)/(((x^2-2x+1)))) is (6x)/((x-1)^3)

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

The first derivative of ((-3x^2)/(((x^2-2x+1)))) is (6x)/((x-1)^3)

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

\frac{6 x}{( x - 1 ) ^{3}}

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

- 3 \cdot \frac{2 x ( x ^{2} - 2 x + 1 ) - ( 2 x - 2 ) x ^{2}}{( x ^{2} - 2 x + 1 ) ^{2}} : \frac{6 x}{( x - 1 ) ^{3}}

= \frac{6 x}{( x - 1 ) ^{3}}

\int_{1}^{\infty} \frac{x}{9 + x ^{2}} d x

\int 1 + tan^{2} (x) d x

x \to 2 lim (\frac{g ( x )}{h ( x )})

\int sin^{6} (θ) d θ

x \to \infty lim (x^{2} - 3 x)

What is the derivative of ((-3x^2)/(((x^2-2x+1)))) ?
The derivative of ((-3x^2)/(((x^2-2x+1)))) is (6x)/((x-1)^3)
What is the first derivative of ((-3x^2)/(((x^2-2x+1)))) ?
The first derivative of ((-3x^2)/(((x^2-2x+1)))) is (6x)/((x-1)^3)