What is the derivative of y=(sqrt(x))/(x+1) ?

The derivative of y=(sqrt(x))/(x+1) is (-x+1)/(2sqrt(x)(x+1)^2)

What is the first derivative of y=(sqrt(x))/(x+1) ?

The first derivative of y=(sqrt(x))/(x+1) is (-x+1)/(2sqrt(x)(x+1)^2)

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derivative of

y = \frac{x}{x + 1}

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

Solution steps

\frac{d}{d x} (\frac{x}{x + 1})

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

\frac{( 3 x ^{2} )}{( 5 - 3 x ) ^{4}}

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

\int (8 + x) d x

\frac{d y}{d x}, x^{2} y + x y^{2} = 6

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