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

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

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

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

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

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

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}

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

Simplify

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

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

(\frac{1 + sin ( t )}{1 - cos ( t )})^{2}

\int_{0}^{\infty} 9 sin^{2} (x) d x

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

x = 6 - y^{2}, x = y

y = x^{3} - 3 x + 2

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