What is the derivative of (sqrt(x)/((9+x))) ?

The derivative of (sqrt(x)/((9+x))) is (9-x)/(2sqrt(x)(9+x)^2)

What is the first derivative of (sqrt(x)/((9+x))) ?

The first derivative of (sqrt(x)/((9+x))) is (9-x)/(2sqrt(x)(9+x)^2)

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\frac{d}{d x} (\frac{x}{( 9 + x )})

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

Solution steps

\frac{d}{d x} (\frac{x}{( 9 + x )})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (9 + x) = 1

= \frac{\frac{1}{2 x} ( 9 + x ) - 1 \cdot x}{( 9 + x ) ^{2}}

Simplify

\frac{\frac{1}{2 x} ( 9 + x ) - 1 \cdot x}{( 9 + x ) ^{2}} : \frac{9 - x}{2 x ( 9 + x ) ^{2}}

= \frac{9 - x}{2 x ( 9 + x ) ^{2}}

\frac{d}{d x} (\frac{6 ^{x} + 5}{6 ^{x} + 2})

\int sin (y) sin (x) d y

\int e^{l n (x + 2)} d x

y^{^{''}} + 1.9 \cdot y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 1

\int \frac{x}{x ^{2} + x - 1} d x

What is the derivative of (sqrt(x)/((9+x))) ?
The derivative of (sqrt(x)/((9+x))) is (9-x)/(2sqrt(x)(9+x)^2)
What is the first derivative of (sqrt(x)/((9+x))) ?
The first derivative of (sqrt(x)/((9+x))) is (9-x)/(2sqrt(x)(9+x)^2)