What is the derivative of (sqrt(x)-6/(sqrt(x)+4)) ?

The derivative of (sqrt(x)-6/(sqrt(x)+4)) is 5/(sqrt(x)(\sqrt{x)+4)^2}

What is the first derivative of (sqrt(x)-6/(sqrt(x)+4)) ?

The first derivative of (sqrt(x)-6/(sqrt(x)+4)) is 5/(sqrt(x)(\sqrt{x)+4)^2}

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\frac{d}{d x} (\frac{x - 6}{x + 4})

\frac{5}{x ( x + 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x - 6}{x + 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x - 6 ) ( x + 4 ) - \frac{d}{d x} ( x + 4 ) ( x - 6 )}{( x + 4 ) ^{2}}

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

\frac{d}{d x} (x + 4) = \frac{1}{2 x}

= \frac{\frac{1}{2 x} ( x + 4 ) - \frac{1}{2 x} ( x - 6 )}{( x + 4 ) ^{2}}

Simplify

\frac{\frac{1}{2 x} ( x + 4 ) - \frac{1}{2 x} ( x - 6 )}{( x + 4 ) ^{2}} : \frac{5}{x ( x + 4 ) ^{2}}

= \frac{5}{x ( x + 4 ) ^{2}}

x \to \infty lim (4 (x - 3))

x \to \infty lim (x (0))

y^{2} - 2 x = 7, y - x = 4

y = 2 sin (π x - y)

\int_{1}^{4} ∣ x - 3 ∣ d x

What is the derivative of (sqrt(x)-6/(sqrt(x)+4)) ?
The derivative of (sqrt(x)-6/(sqrt(x)+4)) is 5/(sqrt(x)(\sqrt{x)+4)^2}
What is the first derivative of (sqrt(x)-6/(sqrt(x)+4)) ?
The first derivative of (sqrt(x)-6/(sqrt(x)+4)) is 5/(sqrt(x)(\sqrt{x)+4)^2}