Differentiate y = arctan(e^(-a/2)) with respect to a, and simplify the result.

Step-by-step solution

1. 1

   Set up the chain rule with an explicit inner variable. Let

   $u=e^{-a/2},\qquad y=\tan^{-1}(u).$

   The derivative of the inverse tangent is $\frac{d}{du}\tan^{-1}u=\frac{1}{1+u^{2}}$, so

   $\frac{dy}{da}=\frac{1}{1+u^{2}}\cdot\frac{du}{da}.$

2. 2

   Differentiate the exponential inner function. The exponent is linear in $a$ with slope $-\tfrac12$, so

   $\frac{du}{da}=-\frac12 e^{-a/2}.$

   The factor $-\tfrac12$ is the whole content of this step; an exponential is its own derivative only when the exponent is exactly $a$.

3. 3

   Square the inner function — this is where the exponents collapse. Since $\left(e^{-a/2}\right)^{2}=e^{-a}$,

   $1+u^{2}=1+e^{-a}.$

   Writing $u^{2}$ as $e^{-a}$ rather than leaving $\left(e^{-a/2}\right)^{2}$ is what makes the final simplification possible; halving the exponent and then squaring is the reason the answer looks so tidy.

4. 4

   Combine into the derivative. Substituting both pieces:

   $\frac{dy}{da}=\frac{-\tfrac12 e^{-a/2}}{1+e^{-a}}=-\frac{e^{-a/2}}{2\left(1+e^{-a}\right)}.$

   The result is negative for every $a$, which makes sense: $e^{-a/2}$ decreases as $a$ grows, and $\tan^{-1}$ is increasing, so the composite must decrease.

5. 5

   Simplify by clearing the negative exponents, then check. Multiplying numerator and denominator by $e^{a}$:

   $\frac{dy}{da}=-\frac{e^{a/2}}{2\left(e^{a}+1\right)}.$

   Both forms agree numerically: at $a=0.8$ a central difference of $y$ gives $-0.23125186$, while each closed form gives $-0.23125186$ ✓. The second form is preferable for large $a$, where $e^{-a}$ underflows.