Question

What is the derivative of $\tanh \left( x \right)$?

Answer 1

Hint: In this question we have to find the derivative of the hyperbolic function. In mathematics, the hyperbolic functions are similar to the trigonometric functions or circular functions. The hyperbolic functions are analogs of the circular function or the trigonometric functions. The basic hyperbolic functions are hyperbolic sine (sinh), hyperbolic cosine (cosh), hyperbolic tangent (tanh).

Complete step by step solution:
Hyperbolic functions work in the same way as the “normal” trigonometric “cuisine” but instead of referring to a unit circle, they refer to a set of hyperbolas. If we compare the derivatives of hyperbolic functions with the derivatives of the standard trigonometric functions, then we find lots of similarities and differences also. For example, the derivatives of the sine functions match:
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\sin x=\cos x \\
 & \Rightarrow \dfrac{d}{dx}\sinh x=\cosh x \\
\end{align}$
But we can see difference also
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\cos x=-\sin x \\
 & \Rightarrow \dfrac{d}{dx}\cosh x=\sinh x \\
\end{align}$
The hyperbolic functions are defined through the algebraic expression that includes the exponential function ${{e}^{x}}$ and its inverse exponential function${{e}^{-x}}$. To find the derivative of the tanhx, we will use the trigonometric rule which is $\dfrac{\sinh x}{\cosh x}=\tanh x$.
Now the value of sinhx in term of exponential function is:
$\Rightarrow \sinh x=\dfrac{1}{2}\left( {{e}^{x}}-{{e}^{-x}} \right)$
Similarly, the value of coshx in term of exponential function is:
$\Rightarrow \cosh x=\dfrac{1}{2}\left( {{e}^{x}}+{{e}^{-x}} \right)$
Now,
$\begin{align}
  & \Rightarrow \tanh x=\dfrac{\sinh x}{\cosh x} \\
 & \Rightarrow \tanh x=\dfrac{\dfrac{1}{2}\left( {{e}^{x}}-{{e}^{-x}} \right)}{\dfrac{1}{2}\left( {{e}^{x}}+{{e}^{-x}} \right)} \\
 & \Rightarrow \tanh x=\dfrac{\left( {{e}^{x}}-{{e}^{-x}} \right)}{\left( {{e}^{x}}+{{e}^{-x}} \right)} \\
\end{align}$
Now we will differentiate it with respect to x. since we can easily see that the above expression will be differentiated by quotient rule.
We know the quotient rule is:
$\Rightarrow \dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{\dfrac{du}{dx}v-\dfrac{dv}{dx}u}{{{v}^{2}}}$
We also know the derivative of ${{e}^{x}}$ is ${{e}^{x}}$, and derivative of ${{e}^{-x}}$ is $-{{e}^{-x}}$ .
Now applying the quotient rule, we get
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \tanh x \right)=\dfrac{{{\left( {{e}^{x}}-{{e}^{-x}} \right)}^{\grave{\ }}}\left( {{e}^{x}}+{{e}^{-x}} \right)-\left( {{e}^{x}}-{{e}^{-x}} \right){{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{\grave{\ }}}}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}} \\
 & \Rightarrow \dfrac{d}{dx}\left( \tanh x \right)=\dfrac{\left( {{e}^{x}}+{{e}^{-x}} \right)\left( {{e}^{x}}+{{e}^{-x}} \right)-\left( {{e}^{x}}-{{e}^{-x}} \right)\left( {{e}^{x}}-{{e}^{-x}} \right)}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}} \\
\end{align}$
Now by more simplifying, we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \tanh x \right)=\dfrac{\left( {{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}-{{\left( {{e}^{x}}-{{e}^{-x}} \right)}^{2}} \right)}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}} \\
 & \Rightarrow \dfrac{d}{dx}\left( \tanh x \right)=1-\dfrac{{{\left( {{e}^{x}}-{{e}^{-x}} \right)}^{2}}}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}} \\
\end{align}\]
Now we can write $\tanh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}$
$\Rightarrow \dfrac{d}{dx}\left( \tanh x \right)=1-{{\tanh }^{2}}x$
Since the hyperbolic function identities are similar to the trigonometric functions, like ${{\cosh }^{2}}x+{{\sinh }^{2}}x=1,1-{{\tanh }^{2}}x={{\operatorname{sech}}^{2}}x,{{\coth }^{2}}x-\cos ec{{h}^{2}}x=1$ .
So, we can also write the derivative of the $\tanh x$ as $\dfrac{d}{dx}\left( \tanh x \right)=1-{{\tanh }^{2}}x={{\operatorname{sech}}^{2}}x$.
Hence we get the derivative of the $\tanh x$ which is ${{\operatorname{sech}}^{2}}x$

Note: The hyperbolic functions are used to define distance in specific non- Euclidean geometry, which means estimating the angles and distances in hyperbolic geometry. The basic difference between trigonometric functions and hyperbolic functions is that trigonometric functions can be defined with the rotations along a circle, while hyperbolic functions can be defined with the use of rotations along a hyperbola.