What is the derivative of (2-x/(sqrt(1-y))) ?

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

What is the first derivative of (2-x/(sqrt(1-y))) ?

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

\int \frac{1}{4} sin (4 x) d x

(- 3, 3), (4, - 5)

\frac{9}{x + 1}

x \to - 1 lim ((x + 2)^{7} (2 x)^{2})

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