Math Doubts

Proof of differentiation of Hyperbolic Tangent function

$x$ 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.

Express Hyperbolic Tangent in Exponential form

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.

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

Simplify the exponential form expression

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the functions in the numerator

The complexity of the function is reduced in the previous step and now, let’s try to simplify the numerator further.


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

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

$= \,\,\,$ $e^{2(x+h)} e^{2x}$ $+$ $e^{2(x+h)}$ $-$ $e^{2x}$ $-$ $1$ $-$ $e^{2x} e^{2(x+h)}$ $-$ $e^{2x}$ $+$ $e^{2(x+h)}$ $+$ $1$

$= \,\,\,$ $e^{2(x+h)} e^{2x}$ $-$ $e^{2x} e^{2(x+h)}$ $+$ $e^{2(x+h)}$ $+$ $e^{2(x+h)}$ $-$ $e^{2x}$ $-$ $e^{2x}$ $-$ $1$ $+$ $1$

Write the mathematical expression in an order for simplifying it further.

$= \,\,\,$ $e^{2(x+h)} e^{2x}$ $-$ $e^{2(x+h)}e^{2x}$ $+$ $e^{2(x+h)}$ $+$ $e^{2(x+h)}$ $-$ $e^{2x}$ $-$ $e^{2x}$ $-$ $1$ $+$ $1$

$= \,\,\,$ $\require{cancel} \cancel{e^{2(x+h)}e^{2x}}$ $-$ $\require{cancel} \cancel{e^{2(x+h)}e^{2x}}$ $+$ $2e^{2(x+h)}$ $-2e^{2x}$ $\require{cancel} -\cancel{1}+\cancel{1}$

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

Continue the derivation of derivative of hyperbolic tangent

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.

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

Separate the constant function

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.

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

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

Find the derivative of tanhx function

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.

Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more