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

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

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

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

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derivative of

f (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}}

\int (\frac{7}{x}) d x

p \to 0 lim (p^{2} sin (\frac{1}{p}))

y^{^{''}} - 18 y^{^{'}} + 117 y = 0

x \to \frac{1}{2} lim (\frac{x + 4}{2 x})

\frac{d}{d x} (\frac{3}{x + 2})

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