Differentiate y = e^x / (e^x + e^-x)^2 and simplify the result.

Step-by-step solution

1. 1

   Rewrite the quotient as a product. Moving the denominator up as a negative power avoids the quotient rule's bulky numerator:

   $y=e^{x}\left(e^{x}+e^{-x}\right)^{-2}$

   Now the product rule $\left(uv\right)'=u'v+uv'$ applies with $u=e^{x}$ and $v=\left(e^{x}+e^{-x}\right)^{-2}$.

2. 2

   Differentiate each piece. For $u$, $u'=e^{x}$. For $v$, the chain rule needs the inside derivative

   $\frac{d}{dx}\left(e^{x}+e^{-x}\right)=e^{x}-e^{-x}$

   (the second term picks up a minus from its own chain rule), so

   $v'=-2\left(e^{x}+e^{-x}\right)^{-3}\left(e^{x}-e^{-x}\right)$

3. 3

   Assemble the product rule.

   $y'=e^{x}\left(e^{x}+e^{-x}\right)^{-2}-2e^{x}\left(e^{x}+e^{-x}\right)^{-3}\left(e^{x}-e^{-x}\right)$

4. 4

   Factor out the lowest power. The common factor is $e^{x}\left(e^{x}+e^{-x}\right)^{-3}$ — take the more negative exponent, so the first term contributes one leftover factor of $\left(e^{x}+e^{-x}\right)$:

   $y'=e^{x}\left(e^{x}+e^{-x}\right)^{-3}\left[\left(e^{x}+e^{-x}\right)-2\left(e^{x}-e^{-x}\right)\right]$

5. 5

   Simplify the bracket and multiply through. Distributing the $-2$:

   $e^{x}+e^{-x}-2e^{x}+2e^{-x}=-e^{x}+3e^{-x}$

   and multiplying by the outside $e^{x}$ uses $e^{x}\cdot e^{-x}=1$:

   $e^{x}\left(-e^{x}+3e^{-x}\right)=-e^{2x}+3$

   giving the compact result

   $\frac{dy}{dx}=\frac{3-e^{2x}}{\left(e^{x}+e^{-x}\right)^{3}}$

6. 6

   Verify numerically and locate the turning point. A central difference with $h=10^{-6}$ matches this formula at $x=-1.2$, $0$, $0.7$ and $1.9$ to within $10^{-6}$ ✓. The derivative vanishes when $e^{2x}=3$, i.e. $x=\tfrac12\ln3\approx0.5493$ — the single maximum of $y$, and the sign change from $+$ to $-$ there confirms it.