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Question: How do you prove that \(\dfrac{d}{dx}\left( \coth x \right)=-{{\operatorname{csch}}^{2}}x\) using th...

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

Explanation

Solution

We use the differentiation of the hyperbolic functions. We use the formulas where ddx(cosh⁡x)=sinh⁡x,ddx(sinh⁡x)=cosh⁡x\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⁡2x−sinh⁡2x=1{{\cosh }^{2}}x-{{\sinh }^{2}}x=1. Then we use the quotient formula of differentiation to find ddx(coth⁡x)=−csch⁡2x\dfrac{d}{dx}\left( \coth x \right)=-{{\operatorname{csch}}^{2}}x where coth⁡x=cosh⁡xsinh⁡x\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 ddx(cosh⁡x)=sinh⁡x\dfrac{d}{dx}\left( \cosh x \right)=\sinh x, ddx(sinh⁡x)=cosh⁡x\dfrac{d}{dx}\left( \sinh x \right)=\cosh x. It is also given that coth⁡x=cosh⁡xsinh⁡x\coth x=\dfrac{\cosh x}{\sinh x}.
The normal trigonometric equation of sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 changes in hyperbolic function as
cosh⁡2x−sinh⁡2x=1{{\cosh }^{2}}x-{{\sinh }^{2}}x=1.
To prove the relation of ddx(coth⁡x)=−csch⁡2x\dfrac{d}{dx}\left( \coth x \right)=-{{\operatorname{csch}}^{2}}x, we will use the quotient rule of differentiation where ddx(uv)=vddx(u)−uddx(v)v2\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,vu,v both are functions of x.
Now for the quotient formula we assume u=cosh⁡x,v=sinh⁡xu=\cosh x,v=\sinh x.
Putting the values, we get
ddx(cosh⁡xsinh⁡x)=sinh⁡xddx(cosh⁡x)−cosh⁡xddx(sinh⁡x)(sinh⁡x)2\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
ddx(cosh⁡xsinh⁡x)=sinh⁡xddx(cosh⁡x)−cosh⁡xddx(sinh⁡x)(sinh⁡x)2 ⇒ddx(coth⁡x)=sinh⁡x(sinh⁡x)−cosh⁡x(cosh⁡x)(sinh⁡x)2 \begin{aligned} & \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{aligned}
We complete the multiplication in the numerator and get
ddx(coth⁡x)=sinh⁡2x−cosh⁡2x(sinh⁡x)2=−cosh⁡2x−sinh⁡2xsinh⁡2x\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⁡2x−sinh⁡2x=1{{\cosh }^{2}}x-{{\sinh }^{2}}x=1 to get
ddx(coth⁡x)=−cosh⁡2x−sinh⁡2xsinh⁡2x=−1sinh⁡2x=−csch⁡2x\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 ddx(coth⁡x)=−csch⁡2x\dfrac{d}{dx}\left( \coth x \right)=-{{\operatorname{csch}}^{2}}x using the definition of coth⁡x=cosh⁡xsinh⁡x\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.