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

The derivative of (sqrt(x)-6/(sqrt(x+6))) is (3(sqrt(x)+1))/(sqrt(x)(x+6)\sqrt{x+6)}

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

The first derivative of (sqrt(x)-6/(sqrt(x+6))) is (3(sqrt(x)+1))/(sqrt(x)(x+6)\sqrt{x+6)}

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

\frac{3 ( x + 1 )}{x ( x + 6 ) x + 6}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{3 ( x + 1 )}{x ( x + 6 ) x + 6}

f (θ) = 13 θ - 8 sin (θ)

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n = 1 \sum \infty 0. 5^{n - 1} \cdot n

\frac{e ^{x}}{2} + \frac{e ^{- x}}{2}

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What is the derivative of (sqrt(x)-6/(sqrt(x+6))) ?
The derivative of (sqrt(x)-6/(sqrt(x+6))) is (3(sqrt(x)+1))/(sqrt(x)(x+6)\sqrt{x+6)}
What is the first derivative of (sqrt(x)-6/(sqrt(x+6))) ?
The first derivative of (sqrt(x)-6/(sqrt(x+6))) is (3(sqrt(x)+1))/(sqrt(x)(x+6)\sqrt{x+6)}