What is the derivative of (1-y/(x-1)) ?

The derivative of (1-y/(x-1)) is (-x\frac{dy)/(dx)+(dy)/(dx)-1+y}{(x-1)^2}

What is the first derivative of (1-y/(x-1)) ?

The first derivative of (1-y/(x-1)) is (-x\frac{dy)/(dx)+(dy)/(dx)-1+y}{(x-1)^2}

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\frac{d}{d x} (\frac{1 - y}{x - 1})

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

Solution steps

\frac{d}{d x} (\frac{1 - y}{x - 1})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (1 - y) = - \frac{d y}{d x}

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

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

Simplify

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

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

\int (cos^{2} (x) cos (x)) d x

\int sin^{2} (t) cos^{3} (t) d t

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

y^{^{'}} = 0.19 y + 200

\int 5 e^{4} d x

What is the derivative of (1-y/(x-1)) ?
The derivative of (1-y/(x-1)) is (-x\frac{dy)/(dx)+(dy)/(dx)-1+y}{(x-1)^2}
What is the first derivative of (1-y/(x-1)) ?
The first derivative of (1-y/(x-1)) is (-x\frac{dy)/(dx)+(dy)/(dx)-1+y}{(x-1)^2}