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

The derivative of (sqrt(x-4))/(sqrt(x+4)) is 4/(sqrt(x-4)(x+4)^{3/2)}

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

The first derivative of (sqrt(x-4))/(sqrt(x+4)) is 4/(sqrt(x-4)(x+4)^{3/2)}

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derivative of

\frac{x - 4}{x + 4}

\frac{4}{x - 4 ( x + 4 ) ^{\frac{3}{2}}}

Solution steps

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

Simplify

\frac{x - 4}{x + 4} : \frac{x - 4}{x + 4}

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x - 4}{x + 4}) = \frac{8}{( x + 4 ) ^{2}}

= \frac{1}{2 \frac{x - 4}{x + 4}} \cdot \frac{8}{( x + 4 ) ^{2}}

Simplify

\frac{1}{2 \frac{x - 4}{x + 4}} \cdot \frac{8}{( x + 4 ) ^{2}} : \frac{4}{x - 4 ( x + 4 ) ^{\frac{3}{2}}}

= \frac{4}{x - 4 ( x + 4 ) ^{\frac{3}{2}}}

y = \frac{1}{3 x}

\frac{d}{d x} (sin (x^{3} - x))

2 y = x, x + 1 = (y - 1)^{2}

\frac{d}{d x} (7 - 8 x)

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

What is the derivative of (sqrt(x-4))/(sqrt(x+4)) ?
The derivative of (sqrt(x-4))/(sqrt(x+4)) is 4/(sqrt(x-4)(x+4)^{3/2)}
What is the first derivative of (sqrt(x-4))/(sqrt(x+4)) ?
The first derivative of (sqrt(x-4))/(sqrt(x+4)) is 4/(sqrt(x-4)(x+4)^{3/2)}