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

The derivative of ln(x+y-ln(x-y)) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}

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

The first derivative of ln(x+y-ln(x-y)) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}

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Popular >

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

\frac{2 x \frac{d y}{d x} - 2 y}{( x + y ) ( x - y )}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (ln (x + y)) = \frac{1 + \frac{d y}{d x}}{x + y}

\frac{d}{d x} (ln (x - y)) = \frac{1 - \frac{d y}{d x}}{x - y}

= \frac{1 + \frac{d y}{d x}}{x + y} - \frac{1 - \frac{d y}{d x}}{x - y}

Simplify

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

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

\frac{\partial}{\partial x} (cos (x) - x cos (y))

\int 5 t^{2} + e^{0.39 t} + 2 d t

\int \frac{t}{a + b t ^{2}} d t

\int 9 e^{- 9 x} d x

sin^{2} (x), sin^{3} (x)

What is the derivative of ln(x+y-ln(x-y)) ?
The derivative of ln(x+y-ln(x-y)) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}
What is the first derivative of ln(x+y-ln(x-y)) ?
The first derivative of ln(x+y-ln(x-y)) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}