Simplifying the derivative of $f(x)= \frac{e^x - e^{-x}}{e^x+e^{-x}}$

Question

I was having some trouble on simplifying the derivative because I didn't know if it's correct. The original function is $$f(x)= \frac{e^x - e^{-x}}{e^x+e^{-x}}$$ What would the simplified derivative be with no negative exponents?

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So far I got $$f'(x)= \frac{(e^x+e^{-x})^2 - (e^x-e^{-x})^2}{(e^x+e^{-x})^2}$$ is this correct?

Answer 1

What you have thus far is correct.

Observe that the numerator of

$$f'(x) = \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$

is a difference of squares. Thus,

\begin{align*} f'(x) & = \frac{[(e^x + e^{-x}) + (e^x - e^{-x})][(e^x + e^{-x}) - (e^x - e^{-x})]}{(e^x + e^{-x})^2}\\ & = \frac{(2e^x)(2e^{-x})}{(e^x + e^{-x})^2}\\ & = \frac{4}{(e^x + e^{-x})^2} \end{align*}

If you do not want any negative exponents, multiply the numerator and denominator by $e^{2x}$ to obtain

\begin{align*} f'(x) & = \frac{4e^{2x}}{(e^x + e^{-x})^2e^{2x}}\\ & = \frac{4e^{2x}}{(e^{2x} + 2 + e^{-2x})e^{2x}}\\ & = \frac{4e^{2x}}{e^{4x} + 2e^{2x} + 1}\\ & = \frac{4e^{2x}}{(e^{2x} + 1)^2} \end{align*}

Alternatively, multiply the numerator and denominator of

$$f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$

by $e^x$ to obtain

$$f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$$

then differentiate to obtain

\begin{align*} f'(x) & = \frac{2e^{2x}(e^{2x} + 1) - 2e^{2x}(e^{2x} - 1)}{(e^{2x} + 1)^2}\\ & = \frac{2^{4x} + 2e^{2x} - 2e^{4x} + 2e^{2x}}{(e^{2x} + 1)^2}\\ & = \frac{4e^{2x}}{(e^{2x} + 1)^2} \end{align*} which agrees with the previous result.

Answer 2

Hint: $f(x)=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=\tanh x \implies f'(x)= \dfrac{1}{\cosh^2 x}$

Answer 3

$y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=\dfrac{e^{2x}-1}{e^{2x}+1}\\ \implies\dfrac{y+1}{y-1}=-e^{2x}\\ \implies (y+1)=e^{2x}(1-y)\\ \implies \dfrac{dy}{dx}=-e^{2x}\dfrac{dy}{dx}+2e^{2x}(1-y)\\\implies \dfrac{dy}{dx}=\dfrac{4e^{2x}}{\left(e^{2x}+1\right)^2}$

Answer 4

First notice that you have a quotient and need the derivative.

Invoke the quotient rule $$\left( \frac{f}{g} \right)^{\prime} = \frac{(e^x + e^{-x})^{\prime} (e^x - e^{-x}) - (e^x - e^{-x})^{\prime}(e^x+e^{-x})}{(e^x + e^{-x})^2}$$

• $(e^x - e^{-x})^{\prime} = e^x - (-1)e^{-x}=e^x + e^{-x} $
• $(e^x + e^{-x})^{\prime} = e^x - e^{-x}$

$$ \Rightarrow \frac{(e^x - e^{-x})(e^x-e^{-x}) - ((e^x +e^{-x})(e^x + e^{-x}))}{(e^x+e^{-x})^2} $$

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

$$\Rightarrow \frac{4}{(e^x + e^{-x})^2}$$

There are various other ways to write the solution:

$\operatorname{sech}^2 x$ as user @GFauxPas stated

There is also $\frac{4}{e^{2x} + 1} - \frac{4}{(e^{2x}+1)^2}$ if you like fractions.

Hope this helps.