What is the derivative of f(x)=(sqrt(x))/(2+x) ?

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

What is the first derivative of f(x)=(sqrt(x))/(2+x) ?

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

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derivative of

f (x) = \frac{x}{2 + x}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{A}{x})

x \to 1 lim (x^{2} + 2 x - \frac{5}{2})

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0, y (0) = 8

\int_{- 1}^{1} e^{y} - (y^{2} - 7) d y

\frac{d}{d x} (lo g_{9} (- (x - 4) (5 x - 2)))

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