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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

= e^{2} \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

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

Simplify

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

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

\frac{d}{d x} (3^{x} sin^{2} (5 x))

\frac{\partial}{\partial x} (8 + x ln (x y - 9))

\frac{\partial}{\partial x} (x cos (x + y) + y)

\int_{- 2}^{0} (x - 2) (x + 1) d x

y = 8 x, y = \frac{5}{2} x, y = 9 - x^{2}

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