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

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

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((x-1)/x)^'

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

\frac{1}{x ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

(x - 1)^{'} = 1

x^{'} = 1

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

Simplify

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

= \frac{1}{x ^{2}}

tan (x), 0, \frac{π}{4}

y = e^{x}, y = e^{- 2 x}, x = ln (4)

10 y^{^{''}} + 20 y^{^{'}} + 6260 y = 0

\frac{d}{d x} (2 x cos (x) - x^{2} sin (x))

\frac{d}{d x} (\frac{5 x + 1}{4 x ^{2} + 1})