What is the derivative of f(x)=(6x)/(6+sqrt(x)) ?

The derivative of f(x)=(6x)/(6+sqrt(x)) is (3(sqrt(x)+12))/((6+sqrt(x))^2)

What is the first derivative of f(x)=(6x)/(6+sqrt(x)) ?

The first derivative of f(x)=(6x)/(6+sqrt(x)) is (3(sqrt(x)+12))/((6+sqrt(x))^2)

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derivative of

f (x) = \frac{6 x}{6 + x}

\frac{3 ( x + 12 )}{( 6 + x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{6 + x})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{6 + x})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d x}{d x} ( 6 + x ) - \frac{d}{d x} ( 6 + x ) x}{( 6 + x ) ^{2}}

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

\frac{d}{d x} (6 + x) = \frac{1}{2 x}

= 6 \cdot \frac{1 \cdot ( 6 + x ) - \frac{1}{2 x} x}{( 6 + x ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( 6 + x ) - \frac{1}{2 x} x}{( 6 + x ) ^{2}} : \frac{3 ( x + 12 )}{( 6 + x ) ^{2}}

= \frac{3 ( x + 12 )}{( 6 + x ) ^{2}}

\int 8 x^{3} - 6 x^{2} + 10 x - 9 d x

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

\frac{d}{d x} (\frac{sec ( x )}{2 + sec ( x )})

3 (x^{2} + y^{2}) d x + x d y - y d x = 0

5 x^{2} - 4 x y + 3 y^{3} = 993

What is the derivative of f(x)=(6x)/(6+sqrt(x)) ?
The derivative of f(x)=(6x)/(6+sqrt(x)) is (3(sqrt(x)+12))/((6+sqrt(x))^2)
What is the first derivative of f(x)=(6x)/(6+sqrt(x)) ?
The first derivative of f(x)=(6x)/(6+sqrt(x)) is (3(sqrt(x)+12))/((6+sqrt(x))^2)