What is the derivative of x+ln(x^2-1) ?

The derivative of x+ln(x^2-1) is 1+(2x)/(x^2-1)

What is the first derivative of x+ln(x^2-1) ?

The first derivative of x+ln(x^2-1) is 1+(2x)/(x^2-1)

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\frac{d}{d x} (x + ln (x^{2} - 1))

1 + \frac{2 x}{x ^{2} - 1}

Solution steps

\frac{d}{d x} (x + ln (x^{2} - 1))

Apply the Sum/Difference Rule:

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

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

\frac{d x}{d x} = 1

\frac{d}{d x} (ln (x^{2} - 1)) = \frac{2 x}{x ^{2} - 1}

= 1 + \frac{2 x}{x ^{2} - 1}

x \to - 2 π lim (sin (x))

h \to 1 lim (\frac{( h - 1 ) ^{3} + 1}{h})

x \to 0 lim (\frac{arcsin ( x )}{x})

\frac{\partial}{\partial y} (sin (y) - tan (x) - y)

ln (x e^{- 2 x})

What is the derivative of x+ln(x^2-1) ?
The derivative of x+ln(x^2-1) is 1+(2x)/(x^2-1)
What is the first derivative of x+ln(x^2-1) ?
The first derivative of x+ln(x^2-1) is 1+(2x)/(x^2-1)