Math Doubts

Hyperbolic tangent function

Formula

$\tanh{x} \,=\, \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

The ratio of subtraction of negative natural exponential function from positive natural exponential function to sum of them is called the hyperbolic tangent function.

Introduction

$x$ is a variable and $e$ is an irrational positive mathematical constant. The positive and negative natural exponential functions are written as $e^x$ and $e^{-x}$ respectively in mathematics.

The subtraction of negative natural exponential function from positive natural exponential function is $e^x\,–\,e^{-x}$. Similarly, the summation of positive and negative natural exponential functions is $e^x+e^{-x}$. The ratio of them is known as hyperbolic tangent in function form.

$\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

The function is in terms of $x$. So, the hyperbolic tan function is usually written as $\tanh{x}$ or $\tanh{(x)}$ in mathematics, and here, $\tanh$ represents hyperbolic tangent.

$\therefore \,\,\, \tanh{x} \,=\, \dfrac{e^x-e^{-x}}{e^x+e^{-x}}$

Note

Remember, the hyperbolic tan function can be defined in terms of any variable, which means there is no rule to write it as $\tanh{(x)}$ always. Therefore, the variable can be defined by any symbol but the hyperbolic tangent function in mathematical form should be written in terms of the corresponding variable.

Examples

$(1) \,\,\,$ $\tanh{(m)} \,=\, \dfrac{e^m-e^{-m}}{e^m+e^{-m}}$

$(2) \,\,\,$ $\tanh{(u)} \,=\, \dfrac{e^u-e^{-u}}{e^u+e^{-u}}$

$(3) \,\,\,$ $\tanh{(y)} \,=\, \dfrac{e^y-e^{-y}}{e^y+e^{-y}}$

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