What is the derivative of y=(sqrt(x))/(8+x) ?

The derivative of y=(sqrt(x))/(8+x) is (8-x)/(2sqrt(x)(8+x)^2)

What is the first derivative of y=(sqrt(x))/(8+x) ?

The first derivative of y=(sqrt(x))/(8+x) is (8-x)/(2sqrt(x)(8+x)^2)

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derivative of

y = \frac{x}{8 + x}

\frac{8 - x}{2 x ( 8 + x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x}{8 + x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ) ( 8 + x ) - \frac{d}{d x} ( 8 + x ) x}{( 8 + x ) ^{2}}

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

\frac{d}{d x} (8 + x) = 1

= \frac{\frac{1}{2 x} ( 8 + x ) - 1 \cdot x}{( 8 + x ) ^{2}}

Simplify

\frac{\frac{1}{2 x} ( 8 + x ) - 1 \cdot x}{( 8 + x ) ^{2}} : \frac{8 - x}{2 x ( 8 + x ) ^{2}}

= \frac{8 - x}{2 x ( 8 + x ) ^{2}}

\frac{d}{d x} (cot^{7} (x))

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 4

\frac{2 x}{( x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (x^{3} + 2 x^{2} - x + 4)

\int \frac{1}{x ^{2} - x ^{3}} d x

What is the derivative of y=(sqrt(x))/(8+x) ?
The derivative of y=(sqrt(x))/(8+x) is (8-x)/(2sqrt(x)(8+x)^2)
What is the first derivative of y=(sqrt(x))/(8+x) ?
The first derivative of y=(sqrt(x))/(8+x) is (8-x)/(2sqrt(x)(8+x)^2)