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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int \frac{1}{x ^{2} + 10 x + 41} d x

(0, - 3), (- 3, 0)

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

\frac{\partial}{\partial x} (x^{a} y^{b})

\frac{\partial}{\partial x} (\frac{( 2 x )}{y})

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