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

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

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

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

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derivative of

y = \frac{x}{x + 3}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (- a e^{- x})

\frac{\partial ^{2}}{\partial y ^{2}} (x^{2} - e^{y^{2}})

\int_{0}^{\frac{π}{8}} x tan^{2} (2 x) d x

\int \frac{1}{4 u ^{2} + 9 - 41 u} d u

- sin (2 t)

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