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

The derivative of f(x)=(2x)/(2+sqrt(x)) is (4+sqrt(x))/((2+sqrt(x))^2)

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

The first derivative of f(x)=(2x)/(2+sqrt(x)) is (4+sqrt(x))/((2+sqrt(x))^2)

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derivative of

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

\frac{4 + x}{( 2 + x ) ^{2}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

2 \cdot \frac{1 \cdot ( 2 + x ) - \frac{1}{2 x} x}{( 2 + x ) ^{2}} : \frac{4 + x}{( 2 + x ) ^{2}}

= \frac{4 + x}{( 2 + x ) ^{2}}

\frac{\partial}{\partial x} (4 x cos (y))

5 sin (4 t)

\frac{\partial}{\partial z} (x^{y})

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0

\int_{1}^{10} \frac{1}{4 x ^{3}} d x

What is the derivative of f(x)=(2x)/(2+sqrt(x)) ?
The derivative of f(x)=(2x)/(2+sqrt(x)) is (4+sqrt(x))/((2+sqrt(x))^2)
What is the first derivative of f(x)=(2x)/(2+sqrt(x)) ?
The first derivative of f(x)=(2x)/(2+sqrt(x)) is (4+sqrt(x))/((2+sqrt(x))^2)