Derivative of tanhx: Formula, Proof | tanhx Derivative

The derivative of tanh(x), denoted by d/dx tanh(x), is equal to sech²x. In this post, we will learn how to differentiate tanh(x), i.e, how to find the derivative of the hyperbolic tan function with respect to x.

The derivative formula of tanh(x) is given as follows:

d/dx[tanh(x)] = sech²x.

How to Differentiate tanh(x)

As tanh x = $\dfrac{\sinh x}{\cosh x}$, by the quotient rule, the derivative of tanh x will be equal to

$(\tanh x)’ =$ $\dfrac{\cosh x \cdot (\sinh x)’ – \sinh x \cdot (\cosh x)’}{{\cosh }^2 x}$ where $f'(x)$ denotes the derivative of f(x).

= $\dfrac{\cosh x \cdot \cosh x – \sinh x \cdot \sinh x}{{\cosh }^2 x}$ as the derivative of cosh(x) is sinh(x) and the derivative of sinh(x) is cosh(x).

= $\dfrac{{\cosh }^2 x – {\sinh }^2 x}{{\cosh }^2 x}$

= $\dfrac{1}{{\cosh }^2 x}$ as we know that cosh²x – sinh²x=1

= sech² x.

Thus the differentiation of tanh x is equal to sech² x.

Derivative of tanh(x) by Quotient Rule

We know that

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

Thus, tanh(x) is a quotient function, so we can use the quotient rule to find its derivative. Now, by the quotient rule of derivative, we have that

$\dfrac{d}{dx}\left(\tanh(x) \right ) =$ $\frac{(e^x+e^{-x})(e^x-e^{-x})’-(e^x-e^{-x})(e^x+e^{-x})’} {(e^x+e^{-x})^2}$.

= $\dfrac{(e^x+e^{-x})(e^x+e^{-x})-(e^x-e^{-x})(e^x-e^{-x})} {(e^x+e^{-x})^2}$ as we know that (e^x)$’$ = e^x and (e^-x)$’$ = -e^-x.

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

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

= $1- \Big[\dfrac{e^x-e^{-x}}{e^x+e^{-x}} \Big]^2$

=$1-\tanh^2 x$ by the definition of tanh(x)

= sech²(x).

So the derivative of tanh(x) is equal to sech²(x), and this is obtained by the quotient rule of derivatives.

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