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

The derivative of y=ln((x-3)/(x+3)) is 6/((x-3)(x+3))

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

The first derivative of y=ln((x-3)/(x+3)) is 6/((x-3)(x+3))

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derivative of

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

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

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

Simplify

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

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

\int (y^{5} - 5 y) d y

e^{s i n (x)}

f (x) = x^{3} + 3 x - 2, at x = 3

\int (x^{1.7} + 9 x^{3.5}) d x

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What is the derivative of y=ln((x-3)/(x+3)) ?
The derivative of y=ln((x-3)/(x+3)) is 6/((x-3)(x+3))
What is the first derivative of y=ln((x-3)/(x+3)) ?
The first derivative of y=ln((x-3)/(x+3)) is 6/((x-3)(x+3))