The derivative of tanh(x), denoted by d/dx tanh(x), is equal to sech^{2}x. 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)] = sech^{2}x.

## 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 cosh^{2}x – sinh^{2}x=1

= sech^{2} x.

Thus the differentiation of tanh x is equal to** **sech^{2} 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)

= sech^{2}(x).

So the derivative of tanh(x) is equal to sech^{2}(x), and this is obtained by the quotient rule of derivatives.

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## FAQs

**Q1: What is the derivative of tanh(x)?**

Answer: The derivative of tanhx is sech^{2}x, that is, d/dx[tanh(x)] = sech^{2}x.

**Q2: If y=tanh(x), then find dy/dx?**

Answer: If y=tanh x, then dy/dx = sech^{2}x.