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

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?
 A: Hint:
$f(x)=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=\tanh x \implies f'(x)= \dfrac{1}{\cosh^2 x}$
A: $y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=\dfrac{e^{2x}-1}{e^{2x}+1}\\ \implies\dfrac{y+1}{y-1}=-e^{2x}\\ \implies (y+1)=e^{2x}(1-y)\\ \implies \dfrac{dy}{dx}=-e^{2x}\dfrac{dy}{dx}+2e^{2x}(1-y)\\\implies \dfrac{dy}{dx}=\dfrac{4e^{2x}}{\left(e^{2x}+1\right)^2}$
A: 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.
A: 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. 
