Question

find the derivative of the following function.
$f(z) = \left(2z + e^{-z^2}\
ight)^6$
$f(z) = \square$

Explanation:

Step1: Apply chain rule

Let $u = 2z + e^{-z^2}$, so $f(z) = u^6$. Then $f'(z) = 6u^5 \cdot u'$.

Step2: Compute $u'$

$u' = \frac{d}{dz}(2z) + \frac{d}{dz}(e^{-z^2}) = 2 + e^{-z^2} \cdot (-2z)$.

Step3: Substitute back $u$

$f'(z) = 6(2z + e^{-z^2})^5 \cdot (2 - 2z e^{-z^2})$.

Answer:

$6(2z + e^{-z^2})^5(2 - 2ze^{-z^2})$