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

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

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

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

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derivative of

y = \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{d}{d x} (\frac{8}{x ^{5}})

(0, 1), (5, - 2)

\int \frac{8 - x}{8 + x} d x

4 x + \frac{6}{7 x}

t \to 0 lim (\frac{1}{t + 1})

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