Maclaurin series for inverse hyperbolics

A question is this type if and only if it asks to find the Maclaurin series expansion for a function involving inverse hyperbolic functions.

5 questions · Challenging +1.3

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Starting from the definition of tanh in terms of exponentials, prove that $\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)$. [3]
Given that $y = \tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)$, show that $( 2 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 1 = 0$.
Hence find the first three terms in the Maclaurin's series for $\tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)$ in the form $$a \ln 3 + b x + c x ^ { 2 } ,$$ where $a , b$ and $c$ are constants to be determined.

Show that $\mathrm { f } ^ { \prime } ( x ) = - \frac { 1 } { 1 + 2 x }$, and find $\mathrm { f } ^ { \prime \prime } ( x )$.
Show that the first three terms of the Maclaurin series for $\mathrm { f } ( x )$ can be written as $\ln a + b x + c x ^ { 2 }$, for constants $a , b$ and $c$ to be found.

Show that $\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 1 } { 2 ( x + 1 ) ^ { 2 } }$.
Hence find the Maclaurin series for $\mathrm { f } ( x )$ up to and including the term in $x ^ { 2 }$.

Using exponentials, prove that $\sinh 2 x = 2 \cosh x \sinh x$.
Hence show that if $\mathrm { f } ( x ) = \sinh ^ { 2 } x$, then $\mathrm { f } ^ { \prime \prime } ( x ) = 2 \cosh 2 x$.
Explain why the coefficients of odd powers in the Maclaurin series for $\sinh ^ { 2 } x$ are all zero.
Find the coefficient of $x ^ { n }$ in this series when $n$ is a positive even number.

Starting from the definitions of $\tanh$ and $\sech$ in terms of exponentials, prove that $$1 - \tanh^2 \theta = \sech^2 \theta.$$ [3]

By differentiating the equation $\tanh y = \cos\left(x + \frac{1}{4}\pi\right)$ with respect to $x$, show that $$\frac{dy}{dx} = -\operatorname{cosec}\left(x + \frac{1}{4}\pi\right).$$ [4]
Hence find the first three terms in the Maclaurin's series for $\tanh^{-1}\left(\cos\left(x + \frac{1}{4}\pi\right)\right)$ in the form $\frac{1}{2}\ln a + bx + cx^2$, giving the exact values of the constants $a$, $b$ and $c$. [5]