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

The derivative of 4/((x^2-2x)^2) is -(8(2x-2))/((x^2-2x)^3)

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

The first derivative of 4/((x^2-2x)^2) is -(8(2x-2))/((x^2-2x)^3)

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derivative of

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

- \frac{8 ( 2 x - 2 )}{( x ^{2} - 2 x ) ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

= 4 (- \frac{2}{( x ^{2} - 2 x ) ^{3}} (2 x - 2))

Simplify

4 (- \frac{2}{( x ^{2} - 2 x ) ^{3}} (2 x - 2)) : - \frac{8 ( 2 x - 2 )}{( x ^{2} - 2 x ) ^{3}}

= - \frac{8 ( 2 x - 2 )}{( x ^{2} - 2 x ) ^{3}}

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

\int cos (3 t) sin (3 t) d t

\int \frac{12 x ^{2}}{4 x ^{3} - 1} d x

\int \frac{4 x + 7}{( 8 x ^{2} + 28 x ) ^{3}} d x

\int 5 csc (x) d x

What is the derivative of 4/((x^2-2x)^2) ?
The derivative of 4/((x^2-2x)^2) is -(8(2x-2))/((x^2-2x)^3)
What is the first derivative of 4/((x^2-2x)^2) ?
The first derivative of 4/((x^2-2x)^2) is -(8(2x-2))/((x^2-2x)^3)