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

The derivative of y=(8x)/(8+sqrt(x)) is (4(sqrt(x)+16))/((8+sqrt(x))^2)

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

The first derivative of y=(8x)/(8+sqrt(x)) is (4(sqrt(x)+16))/((8+sqrt(x))^2)

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derivative of

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

\frac{4 ( x + 16 )}{( 8 + x ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (8 + x) = \frac{1}{2 x}

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

Simplify

8 \cdot \frac{1 \cdot ( 8 + x ) - \frac{1}{2 x} x}{( 8 + x ) ^{2}} : \frac{4 ( x + 16 )}{( 8 + x ) ^{2}}

= \frac{4 ( x + 16 )}{( 8 + x ) ^{2}}

y^{^{''}} - y^{^{'}} = 5 sin (2 x)

\frac{\partial}{\partial x} (yz e^{x z} - 8)

\int 150 x + 7500 d x

\frac{\partial}{\partial y} (2 x + 3)

\int_{0}^{0.6} 8000 x d x

What is the derivative of y=(8x)/(8+sqrt(x)) ?
The derivative of y=(8x)/(8+sqrt(x)) is (4(sqrt(x)+16))/((8+sqrt(x))^2)
What is the first derivative of y=(8x)/(8+sqrt(x)) ?
The first derivative of y=(8x)/(8+sqrt(x)) is (4(sqrt(x)+16))/((8+sqrt(x))^2)