What is the derivative of f(x)=(sqrt(x))/(3+x) ?

The derivative of f(x)=(sqrt(x))/(3+x) is (3-x)/(2sqrt(x)(3+x)^2)

What is the first derivative of f(x)=(sqrt(x))/(3+x) ?

The first derivative of f(x)=(sqrt(x))/(3+x) is (3-x)/(2sqrt(x)(3+x)^2)

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derivative of

f (x) = \frac{x}{3 + x}

\frac{3 - x}{2 x ( 3 + x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x}{3 + x})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (3 + x) = 1

= \frac{\frac{1}{2 x} ( 3 + x ) - 1 \cdot x}{( 3 + x ) ^{2}}

Simplify

\frac{\frac{1}{2 x} ( 3 + x ) - 1 \cdot x}{( 3 + x ) ^{2}} : \frac{3 - x}{2 x ( 3 + x ) ^{2}}

= \frac{3 - x}{2 x ( 3 + x ) ^{2}}

\int_{0}^{π} cos (x) cos (n x) d x

(x^{3} + e^{y}) \frac{d y}{d x} = 3 x^{2}

x \to \infty lim (x - 1)

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} + y = 0

\frac{d}{d x} (\frac{x}{( x - 9 ) ^{2}})

What is the derivative of f(x)=(sqrt(x))/(3+x) ?
The derivative of f(x)=(sqrt(x))/(3+x) is (3-x)/(2sqrt(x)(3+x)^2)
What is the first derivative of f(x)=(sqrt(x))/(3+x) ?
The first derivative of f(x)=(sqrt(x))/(3+x) is (3-x)/(2sqrt(x)(3+x)^2)