What is the (1/(x-2))^' ?

The (1/(x-2))^' is -1/((x-2)^2)

What is the first (1/(x-2))^' ?

The first (1/(x-2))^' is -1/((x-2)^2)

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

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

Solution steps

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

Apply exponent rule:

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

= ((x - 2)^{- 1})^{^{'}}

Apply the chain rule:

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

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

(x - 2)^{'} = 1

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

Simplify

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

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

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

\frac{1 + X}{1 - X}

\int e^{c o s (25 t)} sin (25 t) d t

x + \frac{5}{x}