Find the derivative of Hyperbolic Tangent

derivative of tanh(x)

The derivative of

\tanh(x)

\text{sech}^2(x)

.

Solution. Let

\begin{align*} f(x) = \tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}. \end{align*}

We will use the quotient rule, where we let

\begin{align*} \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{g(x)}{h(x)}, \end{align*}

such that we can determine

\begin{align*} f'(x) = \frac{d}{dx} \tanh(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}. \end{align*}

We know that

\frac{d}{dx} e^x = e^x

and that

\frac{d}{dx} e^{-x} = -e^{-x}

. So we get

\begin{align*} g'(x) = e^x + e^{-x} \quad \text{and} \quad h'(x) = e^x - e^{-x}. \end{align*}

So wrapping everything together, we get

\begin{align*} f'(x) &= \frac{d}{dx} \tanh(x) \\ &= \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} \\ &= \frac{(e^x + e^{-x})(e^x + e^{-x}) - (e^x - e^{-x})(e^x - e^{-x})}{(e^x + e^{-x})^2} \\ &= \frac{e^{2x} + e^{-2x} + 2 - (e^{2x} - e^{-2x} - 2 )}{(e^x + e^{-x})^2} \\ &= \frac{4}{(e^x + e^{-x})^2} \\ &= \frac{2^2}{(e^x + e^{-x})^2} \\ &= \text{sech}^2(x). \end{align*}

So, the derivative of

\tanh(x)

\text{sech}^2(x)