Derivative of Inverse Hyperbolic Cosine of x over a

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Theorem

$\dfrac {\map \d {\map {\cosh^{-1} } {\frac x a} } } {\d x} = \dfrac 1 {\sqrt {x^2 - a^2} }$

where $x > a$.


Corollary 1

$\map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 - a^2} } } = \dfrac 1 {\sqrt {x^2 - a^2} }$

for $x > a$.


Corollary 2

$\map {\dfrac \d {\d x} } {\map \ln {x - \sqrt {x^2 - a^2} } } = -\dfrac 1 {\sqrt {x^2 - a^2} }$

for $x > a$.


Proof

Let $x > a$.

Then $\dfrac x a > 1$ and so:

\(\displaystyle \frac {\map \d {\map {\cosh^{-1} } {\frac x a} } } {\d x}\) \(=\) \(\displaystyle \frac 1 a \frac 1 {\sqrt {\paren {\frac x a}^2} - 1}\) Derivative of $\cosh^{-1}$ and Derivative of Function of Constant Multiple
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \frac 1 {\sqrt {\frac {x^2 - a^2} {a^2} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \frac a {\sqrt {x^2 - a^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt {x^2 - a^2} }\)


$\cosh^{-1} \dfrac x a$ is not defined when $x \le a$.

When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and so $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.

Hence the restriction on the domain.

$\blacksquare$


Also see


Sources