What is the derivative of f(x)=ln((5x+3)/(5x-3)) ?

The derivative of f(x)=ln((5x+3)/(5x-3)) is -(30)/((5x+3)(5x-3))

What is the first derivative of f(x)=ln((5x+3)/(5x-3)) ?

The first derivative of f(x)=ln((5x+3)/(5x-3)) is -(30)/((5x+3)(5x-3))

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derivative of

f (x) = ln (\frac{5 x + 3}{5 x - 3})

- \frac{30}{( 5 x + 3 ) ( 5 x - 3 )}

Solution steps

\frac{d}{d x} (ln (\frac{5 x + 3}{5 x - 3}))

Apply the chain rule:

\frac{1}{\frac{5 x + 3}{5 x - 3}} \frac{d}{d x} (\frac{5 x + 3}{5 x - 3})

= \frac{1}{\frac{5 x + 3}{5 x - 3}} \frac{d}{d x} (\frac{5 x + 3}{5 x - 3})

\frac{d}{d x} (\frac{5 x + 3}{5 x - 3}) = - \frac{30}{( 5 x - 3 ) ^{2}}

= \frac{1}{\frac{5 x + 3}{5 x - 3}} (- \frac{30}{( 5 x - 3 ) ^{2}})

Simplify

\frac{1}{\frac{5 x + 3}{5 x - 3}} (- \frac{30}{( 5 x - 3 ) ^{2}}) : - \frac{30}{( 5 x + 3 ) ( 5 x - 3 )}

= - \frac{30}{( 5 x + 3 ) ( 5 x - 3 )}

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

\frac{\partial}{\partial y} (- 3 y)

y^{^{'}} + \frac{t}{t ^{2} - 16} y = \frac{e ^{t}}{t - 9}

n = 0 \sum \infty cos (nπ)

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

What is the derivative of f(x)=ln((5x+3)/(5x-3)) ?
The derivative of f(x)=ln((5x+3)/(5x-3)) is -(30)/((5x+3)(5x-3))
What is the first derivative of f(x)=ln((5x+3)/(5x-3)) ?
The first derivative of f(x)=ln((5x+3)/(5x-3)) is -(30)/((5x+3)(5x-3))