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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int (4 x^{3} + 1)^{6} 12 x^{2} d x

\int \frac{1 - cos ( 2 x )}{1 + cos ( 2 x )} d x

\int_{0}^{5} 5 - y d y

\frac{t}{t ^{2} + 7}

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