$x$ is a variable and the hyperbolic tangent function is written as $\tanh{(x)}$ mathematically. The derivative of $\tanh{x}$ with respect to $x$ is written as $\dfrac{d}{dx}{\, \tanh{x}}$ in mathematics. The differentiation of hyperbolic tangent can be derived in differential calculus by the definition of a derivative in limit form.
$\dfrac{d}{dx}{\, f{(x)}}$ $\,=\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$
If $f{(x)} \,=\, \tanh{(x)}$, then $f{(x+h)} \,=\, \tanh{(x+h)}$. Substitute them in the definition of a derivative formula to start finding the derivative of hyperbolic tan function.
$\dfrac{d}{dx}{\, \tanh{(x)}}$ $\,=\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\tanh{(x+h)}-\tanh{x}}{h}}$
Now, let’s start deriving the derivative of $\tanh{x}$ function with respect to $x$ in differentical calculus.
According to hyperbolic tangent function, the hyperbolic tangent function can be expressed mathematically in terms of exponential terms in the ratio form.
$\tanh{x} \,=\, \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$
$\tanh{(x+h)} \,=\, \dfrac{e^{(x+h)}-e^{-(x+h)}}{e^{(x+h)}+e^{-(x+h)}}$
Substitute the equivalent values of $\tanh{(x+h)}$ and $\tanh{x}$ functions in the definition of derivative of tan function.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{(x+h)}-e^{-(x+h)}}{e^{(x+h)}+e^{-(x+h)}}-\dfrac{e^x-e^{-x}}{e^x+e^{-x}}}{h}}$
The mathematical expression is completely in complex form. So, let’s try to simplify the whole mathematical expression to move ahead in finding the derivative of $\tanh{x}$ function in calculus.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{(x+h)}-\dfrac{1}{e^{(x+h)}}}{e^{(x+h)}+\dfrac{1}{e^{(x+h)}}}-\dfrac{e^x-\dfrac{1}{e^x}}{e^x+\dfrac{1}{e^x}}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{\dfrac{e^{(x+h)} \times e^{(x+h)} – 1}{e^{(x+h)}}}{\dfrac{e^{(x+h)} \times e^{(x+h)} + 1}{e^{(x+h)}}}-\dfrac{\dfrac{e^x \times e^x – 1}{e^x}}{\dfrac{e^x \times e^x +1}{e^x}}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{\dfrac{{(e^{(x+h)})}^2-1}{e^{(x+h)}}}{\dfrac{{(e^{(x+h)})}^2+1}{e^{(x+h)}}}-\dfrac{\dfrac{{(e^x)}^2-1}{e^x}}{\dfrac{{(e^x)}^2+1}{e^x}}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{\dfrac{e^{2(x+h)}-1}{e^{(x+h)}}}{\dfrac{e^{2(x+h)}+1}{e^{(x+h)}}}-\dfrac{\dfrac{e^{2x}-1}{e^x}}{\dfrac{e^{2x}+1}{e^x}}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{2(x+h)}-1}{e^{2(x+h)}+1} \times \dfrac{e^{(x+h)}}{e^{(x+h)}}-\dfrac{e^{2x}-1}{e^{2x}+1} \times \dfrac{e^{x}}{e^{x}}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \require{cancel} \dfrac{\dfrac{e^{2(x+h)}-1}{e^{2(x+h)}+1} \times \dfrac{\cancel{e^{(x+h)}}}{\cancel{e^{(x+h)}}}-\dfrac{e^{2x}-1}{e^{2x}+1} \times \dfrac{\cancel{e^{x}}}{\cancel{e^{x}}}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{2(x+h)}-1}{e^{2(x+h)}+1} \times 1-\dfrac{e^{2x}-1}{e^{2x}+1} \times 1}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{2(x+h)}-1}{e^{2(x+h)}+1}-\dfrac{e^{2x}-1}{e^{2x}+1}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{{(e^{2(x+h)}-1)}{(e^{2x}+1)}-{(e^{2x}-1)}{(e^{2(x+h)}+1)}}{{(e^{2(x+h)}+1)}{(e^{2x}+1)}}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{{(e^{2(x+h)}-1)}{(e^{2x}+1)}-{(e^{2x}-1)}{(e^{2(x+h)}+1)}}{{(e^{2(x+h)}+1)}{(e^{2x}+1)}}}{\dfrac{h}{1}}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{{(e^{2(x+h)}-1)}{(e^{2x}+1)}-{(e^{2x}-1)}{(e^{2(x+h)}+1)}}{{(e^{2(x+h)}+1)}{(e^{2x}+1)}} \times \dfrac{1}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\Big[{(e^{2(x+h)}-1)}{(e^{2x}+1)}-{(e^{2x}-1)}{(e^{2(x+h)}+1)}\Big] \times 1}{{(e^{2(x+h)}+1)}{(e^{2x}+1)} \times h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{{(e^{2(x+h)}-1)}{(e^{2x}+1)}-{(e^{2x}-1)}{(e^{2(x+h)}+1)}}{h{(e^{2(x+h)}+1)}{(e^{2x}+1)}}}$
The complexity of the function is reduced in the previous step and now, let’s try to simplify the numerator further.
${(e^{2(x+h)}-1)}{(e^{2x}+1)}-{(e^{2x}-1)}{(e^{2(x+h)}+1)}$
$= \,\,\,$ $e^{2(x+h)} \times e^{2x}$ $+$ $e^{2(x+h)} \times 1$ $-1 \times e^{2x}$ $-1 \times 1$ $-$ $\Big($ $e^{2x} \times e^{2(x+h)}$ $+$ $e^{2x} \times 1$ $-1 \times e^{2(x+h)}$ $-1 \times 1$ $\Big)$
$= \,\,\,$ $e^{2(x+h)} e^{2x}$ $+$ $e^{2(x+h)}$ $-$ $e^{2x}$ $-$ $1$ $-$ $\Big($ $e^{2x} e^{2(x+h)}$ $+$ $e^{2x}$ $-$ $e^{2(x+h)}$ $-$ $1$ $\Big)$
$= \,\,\,$ $e^{2(x+h)} e^{2x}$ $+$ $e^{2(x+h)}$ $-$ $e^{2x}$ $-$ $1$ $-$ $e^{2x} e^{2(x+h)}$ $-$ $e^{2x}$ $+$ $e^{2(x+h)}$ $+$ $1$
$= \,\,\,$ $e^{2(x+h)} e^{2x}$ $-$ $e^{2x} e^{2(x+h)}$ $+$ $e^{2(x+h)}$ $+$ $e^{2(x+h)}$ $-$ $e^{2x}$ $-$ $e^{2x}$ $-$ $1$ $+$ $1$
Write the mathematical expression in an order for simplifying it further.
$= \,\,\,$ $e^{2(x+h)} e^{2x}$ $-$ $e^{2(x+h)}e^{2x}$ $+$ $e^{2(x+h)}$ $+$ $e^{2(x+h)}$ $-$ $e^{2x}$ $-$ $e^{2x}$ $-$ $1$ $+$ $1$
$= \,\,\,$ $\require{cancel} \cancel{e^{2(x+h)}e^{2x}}$ $-$ $\require{cancel} \cancel{e^{2(x+h)}e^{2x}}$ $+$ $2e^{2(x+h)}$ $-2e^{2x}$ $\require{cancel} -\cancel{1}+\cancel{1}$
$= \,\,\,$ $2e^{2(x+h)}-2e^{2x}$
The value of the ${(e^{2(x+h)}-1)}{(e^{2x}+1)}-{(e^{2x}-1)}{(e^{2(x+h)}+1)}$ in the numerator can be replaced by its simplified function form $2e^{2(x+h)}$ $-2e^{2x}$.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{2e^{2(x+h)}-2e^{2x}}{h{(e^{2(x+h)}+1)}{(e^{2x}+1)}}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{2e^{2x+2h}-2e^{2x}}{h{(e^{2(x+h)}+1)}{(e^{2x}+1)}}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{2e^{2x}e^{2h}-2e^{2x}}{h{(e^{2(x+h)}+1)}{(e^{2x}+1)}}}$
The factor $e^{2x}$ is common in both terms in the numerator and it can be taken common from them.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{2e^{2x}{(e^{2h}-1)}}{h{(e^{2(x+h)}+1)}{(e^{2x}+1)}}}$
The limit of the function is actually defined in terms of $h$ and the function in terms of the $x$ is a constant function in this case. So, try to separate the constant function from the whole function.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \Bigg[\dfrac{2e^{2x}}{e^{2x}+1} \times \dfrac{e^{2h}-1}{h{(e^{2(x+h)}+1)}} \Bigg]}$
$= \,\,\,$ $\dfrac{2e^{2x}}{e^{2x}+1} \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{2h}-1}{h{(e^{2(x+h)}+1)}} }$
Now, split the function in terms of $h$ into two multiplying factors for our convenience. It plays a vital rule in getting derivative of hyperbolic tangent in few steps.
$= \,\,\,$ $\dfrac{2e^{2x}}{e^{2x}+1} \large \displaystyle \lim_{h \to 0}{\normalsize \Bigg[ \dfrac{e^{2h}-1}{h} \times \dfrac{1}{e^{2(x+h)}+1} \Bigg] }$
According to multiplication rule of limits, the limit rule is applied to both multiplying functions.
$= \,\,\,$ $\dfrac{2e^{2x}}{e^{2x}+1}\Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{2h}-1}{h}} \times \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{1}{e^{2(x+h)}+1} \Bigg] }$
Now, substitute $h$ is equal to zero in the second multiplying factor and do not disturb the first limit function at this time.
$= \,\,\,$ $\dfrac{2e^{2x}}{e^{2x}+1}\Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{2h}-1}{h}} \times \normalsize \displaystyle \dfrac{1}{e^{2(x+0)}+1} \Bigg] $
$= \,\,\,$ $\dfrac{2e^{2x}}{e^{2x}+1}\Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{2h}-1}{h}} \times \normalsize \displaystyle \dfrac{1}{e^{2x}+1} \Bigg] $
Separate the constant function once again and it helps us to simplify the expression easily.
$= \,\,\,$ $\dfrac{2e^{2x}}{e^{2x}+1} \times \dfrac{1}{e^{2x}+1} \Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{2h}-1}{h}} \normalsize \Bigg]$
$= \,\,\,$ $\dfrac{2e^{2x} \times 1}{(e^{2x}+1) \times (e^{2x}+1)} \Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{2h}-1}{h}} \normalsize \Bigg]$
$= \,\,\,$ $\dfrac{2e^{2x}}{{(e^{2x}+1)}^2} \Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{2h}-1}{h}} \normalsize \Bigg]$
There is a limit rule in terms of exponential function but the denominator should same as the exponent of the exponential term.
$= \,\,\,$ $\dfrac{2e^{2x}}{{(e^{2x}+1)}^2} \Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{2 \times (e^{2h}-1)}{2 \times h}} \normalsize \Bigg]$
$= \,\,\,$ $\dfrac{2e^{2x}}{{(e^{2x}+1)}^2} \Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{2(e^{2h}-1)}{2h}} \normalsize \Bigg]$
$= \,\,\,$ $\dfrac{2e^{2x} \times 2}{{(e^{2x}+1)}^2} \Bigg[ \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{2h}-1}{2h}} \normalsize \Bigg]$
If $h \to 0$ then $2h \to 2 \times 0$. Therefore, the value of $2h$ also approaches $0$. Now, write $2h \to 0$ instead of $h \to 0$.
$= \,\,\,$ $\dfrac{4e^{2x}}{{(e^{2x}+1)}^2} \Bigg[ \large \displaystyle \lim_{2h \to 0}{\normalsize \dfrac{e^{2h}-1}{2h}} \normalsize \Bigg]$
As per the exponential limit rule, the value of $\dfrac{e^{2h}-1}{2h}$ as $2h$ tends to $0$ is equal to one.
$= \,\,\,$ $\dfrac{4e^{2x}}{{(e^{2x}+1)}^2} \Big[ 1 \Big]$
$= \,\,\,$ $\dfrac{4e^{2x}}{{(e^{2x}+1)}^2} \times 1$
It is time to simplify the exponential function to get the differentiation of hyperbolic tangent in differential calculus.
$= \,\,\,$ $\dfrac{4e^{2x}}{{(e^{2x}+1)}^2}$
$= \,\,\,$ $\dfrac{2^2{(e^{x})}^2}{{(e^{2x}+1)}^2}$
$= \,\,\,$ $\dfrac{{(2e^{x})}^2}{{(e^{2x}+1)}^2}$
$= \,\,\,$ ${\Bigg(\dfrac{2e^{x}}{e^{2x}+1}\Bigg)}^2$
$= \,\,\,$ ${\Bigg( \dfrac{2}{ \dfrac{e^{2x}+1}{e^x} } \Bigg)}^2$
$= \,\,\,$ ${\Bigg( \dfrac{2}{ \dfrac{e^{2x}}{e^x}+\dfrac{1}{e^x} } \Bigg)}^2$
$= \,\,\,$ ${\Bigg( \dfrac{2}{e^{2x} \times e^{-x} + e^{-x} } \Bigg)}^2$
$= \,\,\,$ ${\Bigg( \dfrac{2}{e^{2x-x}+ e^{-x}}\Bigg)}^2$
$= \,\,\,$ ${\Bigg( \dfrac{2}{e^{x}+ e^{-x}}\Bigg)}^2$
As per the definition of hyperbolic secant, the function in exponential form is equal to $\operatorname{sech}{x}$.
$= \,\,\,$ ${(\operatorname{sech}{x})}^2$
$\therefore \,\,\,\,\,\, \dfrac{d}{dx}{\, \tanh{x}}$ $\,=\,$ $\operatorname{sech^2}{x}$
Therefore, it is proved that the derivative or differentiation of hyperbolic tangent is equal to square of hyperbolic secant.
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