# Proof of Derivative of Hyperbolic Tangent function

The hyperbolic tangent function is written as $\tanh{x}$ when $x$ is used to denote a variable. The differentiation or the derivative of hyperbolic tan function with respect to $x$ is expressed mathematically as follows.

$\dfrac{d}{dx}{\, \Big(\tanh{(x)}\Big)}$

The derivative formula of the hyperbolic tan function can be proved mathematically in differential calculus by the first principle of the differentiation. Therefore, let’s learn the step by step process to derive the differentiation rule of hyperbolic tangent function.

### Derivative of Hyperbolic Tan function in Limit form

The derivative formula of hyperbolic tangent function can be derived in limit form by the fundamental definition of the derivative in differential calculus.

$\dfrac{d}{dx}{\, (\tanh{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\tanh{(x+\Delta x)}-\tanh{x}}{\Delta x}}$

When $\Delta x$ is simply represented by $h$, then the expression in the right hand side of the equation can be written in $h$ instead of $\Delta x$.

$\implies$ $\dfrac{d}{dx}{\, (\tanh{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tanh{(x+h)}-\tanh{x}}{h}}$

The differentiation of $\tanh{(x)}$ function with respect to $x$ can be proved in differential calculus from the first principle of differentiation.

### Evaluate the Limit by the Direct Substitution

First of all, let’s try the direct substitution method for evaluating the derivative of the hyperbolic tangent function as $h$ approaches zero.

$= \,\,\,$ $\dfrac{\tanh{(x+0)}-\tanh{x}}{0}$

$= \,\,\,$ $\dfrac{\tanh{x}-\tanh{x}}{0}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\tanh{x}}-\cancel{\tanh{x}}}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

The direct substitution method clears that the derivative of the $\tanh{x}$ function is indeterminate but it is not true. Hence, we have to derive the differentiation of the hyperbolic tan function in a different approach.

### Simplify the Hyperbolic Cotangent function

Now, focus on the expression of the derivative of hyperbolic tan function to evaluate it in another method.

$\implies$ $\dfrac{d}{dx}{\, (\tanh{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\tanh{(x+h)}-\tanh{x}}{h}}$

Each hyperbolic tangent function can be expressed in terms of natural exponential functions.

$\implies$ $\dfrac{d}{dx}{\, (\tanh{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{ \dfrac{e^(x+h)}-e^-(x+h)}}{e^(x+h)}+e^-(x+h)}}-\dfrac{e^x}-e^-x}}{e^x}+e^-x}} }{h}$

Let us simplify this mathematical expression firstly before preparing for evaluating the differentiation of hyperbolic tan function.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{ \dfrac{\Big(e^(x+h)}-e^-(x+h)}\Big)\Big(e^x}+e^-x}\Big) -\Big(e^x}-e^-x}\Big)\Big(e^(x+h)}+e^-(x+h)}\Big) }{\Big(e^(x+h)}+e^-(x+h)}\Big)\Big(e^x}+e^-x}\Big) } }{h}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^(x+h)}-e^-(x+h)}\Big)\Big(e^x}+e^-x}\Big) -\Big(e^x}-e^-x}\Big)\Big(e^(x+h)}+e^-(x+h)}\Big) }{\Big(e^(x+h)}+e^-(x+h)}\Big)\Big(e^x}+e^-x}\Big) \times h}$

The below mathematical expression is an expression in the numerator of the above expression.

$\Big(e^(x+h)}-e^-(x+h)}\Big) \Big(e^x}+e^-x}\Big$ $-$ $\Big(e^x}-e^-x}\Big) \Big(e^(x+h)}+e^-(x+h)}\Big$

For evaluting the differentiation of hyperbolic tan function, we have to simplify it firstly.

$=\,\,\,$ $\Big($ $e^(x+h)} \times e^x$ $+$ $e^(x+h)} \times e^-x$ $-$ $e^-(x+h)} \times e^x$ $-$ $e^-(x+h)} \times e^-x$ $\Big)$ $-$ $\Big(e^x} \times e^(x+h)$ $+$ $e^x} \times e^-(x+h)$ $-$ $e^-x} \times e^(x+h)$ $-$ $e^-x} \times e^-(x+h)$ $\Big)$

$=\,\,\,$ $\Big($ $e^(x+h)+x$ $+$ $e^(x+h)-x$ $-$ $e^-(x+h)+x$ $-$ $e^-(x+h)-x$ $\Big)$ $-$ $\Big(e^x+(x+h)$ $+$ $e^x-(x+h)$ $-$ $e^-x+(x+h)$ $-$ $e^-x-(x+h)$ $\Big)$

$=\,\,\,$ $\Big($ $e^x+h+x$ $+$ $e^x+h-x$ $-$ $e^-x-h+x$ $-$ $e^-x-h-x$ $\Big)$ $-$ $\Big(e^x+x+h$ $+$ $e^x-x-h$ $-$ $e^-x+x+h$ $-$ $e^-x-x-h$ $\Big)$

$=\,\,\,$ $\Big($ $e^x+x+h$ $+$ $e^x-x+h$ $-$ $e^-x+x-h$ $-$ $e^-x-x-h$ $\Big)$ $-$ $\Big(e^x+x+h$ $+$ $e^x-x-h$ $-$ $e^-x+x+h$ $-$ $e^-x-x-h$ $\Big)$

$=\,\,\,$ $\require{cancel} \Big($ $e^2x+h$ $+$ $e^\cancel{x}-\cancel{x}+h$ $-$ $e^-\cancel{x}+\cancel{x}-h$ $-$ $e^-2x-h$ $\Big)$ $-$ $\Big(e^2x+h$ $+$ $e^\cancel{x}-\cancel{x}-h$ $-$ $e^-\cancel{x}+\cancel{x}+h$ $-$ $e^-2x-h$ $\Big)$

$=\,\,\,$ $\Big($ $e^2x+h$ $+$ $e^h$ $-$ $e^-h$ $-$ $e^-2x-h$ $\Big)$ $-$ $\Big(e^2x+h$ $+$ $e^-h$ $-$ $e^h$ $-$ $e^-2x-h$ $\Big)$

$=\,\,\,$ $e^2x+h$ $+$ $e^h$ $-$ $e^-h$ $-$ $e^-2x-h$ $-$ $e^2x+h$ $-$ $e^-h$ $+$ $e^h$ $+$ $e^-2x-h$

$=\,\,\,$ $e^2x+h$ $-$ $e^2x+h$ $+$ $e^h$ $+$ $e^h$ $-$ $e^-h$ $-$ $e^-h$ $-$ $e^-2x-h$ $+$ $e^-2x-h$

$=\,\,\,$ $\cancel{e^2x+h}$ $-$ $\cancel{e^2x+h}$ $+$ $2e^h$ $-$ $2e^-h$ $-$ $\cancel{e^-2x-h}$ $+$ $\cancel{e^-2x-h}$

$=\,\,\,$ $2e^h}-2e^-h$

$=\,\,\,$ $2\Big(e^h}-e^-h}\Big$

It is a simplified form of the numerator. So, it can be replaced in the numerator for moving to next step in proving the differentiation of the hyperbolic tan function from first principle.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\Big(e^h}-e^-h}\Big) }{\Big(e^(x+h)}+e^-(x+h)}\Big)\Big(e^x}+e^-x}\Big) \times h}$

### Evaluate the Limit of the function

If the direct substitution method is used to evaluate the limit of the function, the limit is equal to indeterminate. Hence, the function must be factored as product of the functions to evaluate the limit of the function.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[ \dfrac{2}{\Big(e^(x+h)}+e^-(x+h)}\Big)\Big(e^x}+e^-x}\Big)}$ $\times$ $\dfrac{\Big(e^h}-e^-h}\Big)}{h} \Bigg$

We can use the product rule of the limits to find the limit of the product of functions by the product of their limits.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2}{\Big(e^(x+h)}+e^-(x+h)}\Big)\Big(e^x}+e^-x}\Big)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^h}-e^-h}\Big)}{h}$

The limit of the first function as $h$ approaches $0$ can be evaluated by the direct substitution method but leave the second factor at this time.

$=\,\,\,$ $\dfrac{2}{\Big(e^(x+0)}+e^-(x+0)}\Big)\Big(e^x}+e^-x}\Big)$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^h}-e^-h}\Big)}{h}$

$=\,\,\,$ $\dfrac{2}{\Big(e^(x)}+e^-(x)}\Big)\Big(e^x}+e^-x}\Big)$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^h}-e^-h}\Big)}{h}$

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)\Big(e^x}+e^-x}\Big)$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^h}-e^-h}\Big)}{h}$

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^h}-e^-h}\Big)}{h}$

Now, we can focus on evaluating the second function.

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^h}-e^-h}+1-1\Big)}{h}$

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^h}-1-e^-h}+1\Big)}{h}$

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{e^h}-1-e^-h}+1}{h}\Bigg)$

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{e^h}-1}{h}+\dfrac{-e^-h}+1}{h}\Bigg)$

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{e^h}-1}{h}+\dfrac{-(e^-h}-1)}{h}\Bigg)$

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{e^h}-1}{h}+\dfrac{e^-h}-1}{-h}\Bigg)$

It can be simplified by the sum rule of limits.

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^h}-1}{h}$ $+$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^-h}-1}{-h}\Bigg)$

When $h \,\to\, 0$, then $-h \,\to\, 0$. Therefore, $-h$ approaches $0$ when $h$ approaches $0$.

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^h}-1}{h}$ $+$ $\displaystyle \large \lim_{-h \,\to\, 0}{\normalsize \dfrac{e^-h}-1}{-h}\Bigg)$

The limit of (e^x-1)/x as x approaches 0 formula can be used to evaluate the limit of each function.

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\Big(1+1\Big)$

$=\,\,\,$ $\dfrac{2}{\Big(e^x}+e^-x}\Big)^2$ $\times$ $\Big(2\Big)$

$=\,\,\,$ $\dfrac{(2 \times 2)}{\Big(e^x}+e^-x}\Big)^2$

$=\,\,\,$ $\dfrac{2^2}{\Big(e^x}+e^-x}\Big)^2$

$=\,\,\,$ $\Bigg(\dfrac{2}{e^x}+e^-x}}\Bigg)^$

The function in terms of natural exponential functions can be expressed as the hyperbolic secant function simply.

$=\,\,\,$ $\Big(\operatorname{sech}{x}\Big)^2$

$\therefore\,\,\,$ $\dfrac{d}{dx}{\, (\tanh{x})}$ $\,=\,$ $\operatorname{sech}^2{x}$

Thus, the derivative of hyperbolic tangent function is derived in differential calculus by the first principle of differentiation.

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