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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = x in x

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

\frac{d}{d x} (c e^{\frac{y}{x}})

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

\frac{d}{d x} (5 x^{2 π})

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