What is the derivative of y=ln(sqrt((x+3)/(x-3))) ?

The derivative of y=ln(sqrt((x+3)/(x-3))) is -3/((x-3)(x+3))

What is the first derivative of y=ln(sqrt((x+3)/(x-3))) ?

The first derivative of y=ln(sqrt((x+3)/(x-3))) is -3/((x-3)(x+3))

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derivative of

y = ln (\frac{x + 3}{x - 3})

- \frac{3}{( x - 3 ) ( x + 3 )}

Solution steps

\frac{d}{d x} (ln (\frac{x + 3}{x - 3}))

Apply the chain rule:

\frac{1}{\frac{x + 3}{x - 3}} \frac{d}{d x} (\frac{x + 3}{x - 3})

= \frac{1}{\frac{x + 3}{x - 3}} \frac{d}{d x} (\frac{x + 3}{x - 3})

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

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

Simplify

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

= - \frac{3}{( x - 3 ) ( x + 3 )}

\frac{\partial}{\partial x} (y^{6} - 5 x y)

\int 1 + x^{2} d x

\frac{\partial}{\partial z} (\frac{1}{x y z ^{2}})

\frac{d y}{d x} = \frac{- x y ^{2}}{2}

x \to 2 lim ((x - 2) (2^{x} - 4))

What is the derivative of y=ln(sqrt((x+3)/(x-3))) ?
The derivative of y=ln(sqrt((x+3)/(x-3))) is -3/((x-3)(x+3))
What is the first derivative of y=ln(sqrt((x+3)/(x-3))) ?
The first derivative of y=ln(sqrt((x+3)/(x-3))) is -3/((x-3)(x+3))