What is the derivative of y=ln(sqrt((a+bx)/(a-bx))) ?

The derivative of y=ln(sqrt((a+bx)/(a-bx))) is (ab)/((-bx+a)(bx+a))

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

The first derivative of y=ln(sqrt((a+bx)/(a-bx))) is (ab)/((-bx+a)(bx+a))

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derivative of

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

\frac{ab}{( - b x + a ) ( b x + a )}

Solution steps

\frac{d}{d x} (ln (\frac{a + b x}{a - b x}))

Apply the chain rule:

\frac{1}{\frac{a + b x}{a - b x}} \frac{d}{d x} (\frac{a + b x}{a - b x})

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

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

= \frac{1}{\frac{a + b x}{a - b x}} \cdot \frac{ab}{b x + a ( - b x + a ) ^{\frac{3}{2}}}

Simplify

\frac{1}{\frac{a + b x}{a - b x}} \cdot \frac{ab}{b x + a ( - b x + a ) ^{\frac{3}{2}}} : \frac{ab}{( - b x + a ) ( b x + a )}

= \frac{ab}{( - b x + a ) ( b x + a )}

cos (x + 1)

\frac{\partial}{\partial x} (12 y^{5} x^{3} + 35 y^{8} x^{6})

6 (t + 1) \frac{d y}{d t} - 5 y = 5 t

x \to 0 lim (\frac{( x ^{2} - 2 )}{x})

\frac{d}{d x} (x^{4} + y^{4} + z^{4} - 4 x y z)

What is the derivative of y=ln(sqrt((a+bx)/(a-bx))) ?
The derivative of y=ln(sqrt((a+bx)/(a-bx))) is (ab)/((-bx+a)(bx+a))
What is the first derivative of y=ln(sqrt((a+bx)/(a-bx))) ?
The first derivative of y=ln(sqrt((a+bx)/(a-bx))) is (ab)/((-bx+a)(bx+a))