Question

How do you prove that $\dfrac{d}{dx}\left( \coth x \right)=-{{\operatorname{csch}}^{2}}x$ using the definition of $\coth x=\dfrac{\cosh x}{\sinh x}$?

Answer 1

Hint: We use the differentiation of the hyperbolic functions. We use the formulas where $\dfrac{d}{dx}\left( \cosh x \right)=\sinh x,\dfrac{d}{dx}\left( \sinh x \right)=\cosh x$. We also use the identity formula of ${{\cosh }^{2}}x-{{\sinh }^{2}}x=1$. Then we use the quotient formula of differentiation to find $\dfrac{d}{dx}\left( \coth x \right)=-{{\operatorname{csch}}^{2}}x$ where $\coth x=\dfrac{\cosh x}{\sinh x}$.

Complete step by step solution:
The given differentiation is the hyperbolic version of normal differentiation.
The basic differentiations follow same differentiation rules where $\dfrac{d}{dx}\left( \cosh x \right)=\sinh x$, $\dfrac{d}{dx}\left( \sinh x \right)=\cosh x$. It is also given that $\coth x=\dfrac{\cosh x}{\sinh x}$.
The normal trigonometric equation of ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ changes in hyperbolic function as
${{\cosh }^{2}}x-{{\sinh }^{2}}x=1$.
To prove the relation of $\dfrac{d}{dx}\left( \coth x \right)=-{{\operatorname{csch}}^{2}}x$, we will use the quotient rule of differentiation where $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}\left( u \right)-u\dfrac{d}{dx}\left( v \right)}{{{v}^{2}}}$. Here $u,v$ both are functions of x.
Now for the quotient formula we assume $u=\cosh x,v=\sinh x$.
Putting the values, we get
$\dfrac{d}{dx}\left( \dfrac{\cosh x}{\sinh x} \right)=\dfrac{\sinh x\dfrac{d}{dx}\left( \cosh x \right)-\cosh x\dfrac{d}{dx}\left( \sinh x \right)}{{{\left( \sinh x \right)}^{2}}}$.
We now put the differentiated values and get
$\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{\cosh x}{\sinh x} \right)=\dfrac{\sinh x\dfrac{d}{dx}\left( \cosh x \right)-\cosh x\dfrac{d}{dx}\left( \sinh x \right)}{{{\left( \sinh x \right)}^{2}}} \\
 & \Rightarrow \dfrac{d}{dx}\left( \coth x \right)=\dfrac{\sinh x\left( \sinh x \right)-\cosh x\left( \cosh x \right)}{{{\left( \sinh x \right)}^{2}}} \\
\end{align}$
We complete the multiplication in the numerator and get
$\dfrac{d}{dx}\left( \coth x \right)=\dfrac{{{\sinh }^{2}}x-{{\cosh }^{2}}x}{{{\left( \sinh x \right)}^{2}}}=-\dfrac{{{\cosh }^{2}}x-{{\sinh }^{2}}x}{{{\sinh }^{2}}x}$.
Now again we put the values ${{\cosh }^{2}}x-{{\sinh }^{2}}x=1$ to get
$\dfrac{d}{dx}\left( \coth x \right)=-\dfrac{{{\cosh }^{2}}x-{{\sinh }^{2}}x}{{{\sinh }^{2}}x}=-\dfrac{1}{{{\sinh }^{2}}x}=-{{\operatorname{csch}}^{2}}x$.
Thus, we proved that $\dfrac{d}{dx}\left( \coth x \right)=-{{\operatorname{csch}}^{2}}x$ using the definition of $\coth x=\dfrac{\cosh x}{\sinh x}$.

Note: Hyperbolic functions are related to the natural exponential function as well the circular sine and cosine functions. They are called “hyperbolic” because the relationship between the sine and cosine functions is the same as a unit hyperbola.