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

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

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

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

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\frac{d}{d x} (\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}}

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

\frac{\partial}{\partial x} (e^{- 5 t} cos (π x))

x \to - 1 lim (- 1)

\int 2 x (x^{2} + 13)^{7} d x

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