I am trying to obtain the derivative of the following expression with respect to $x$, but not getting it correctly.
$$F = \biggl\{\frac{\exp(\delta){\sigma_5}}{\sqrt{\alpha^2\sigma^2_2\sigma^2_4}}\biggl[\sum_{d=0}^{\infty}W_{-d-0.5,0}(2\delta)\cdot\gamma(1+d,\frac{x}{A_1\sigma^2_5})\biggr]\biggr\}^M$$
where $W_{.,.}(.) $ is Whittaker function, $\gamma(.,.)$ is lower incomplete Gamma function and $x$ is differentiation variable and all others are constant.
Maybe someone has encountered similar derivative before. I am thankful for any advice on how to solve it.
To find the derivative of a "complicated function" divide it into parts, using the sum, product, and chain rules.
$F(x)= (u(x))^m$ where $u(x)= \frac{exp(\delta)}{\sqrt{\alpha^2\sigma_2^2\sigma_4^2}}v(x)$. $v(x)= \sum_{d= 0}^\infty W_{-d-0.5,0}(2\delta)\cdot \gamma(1+ d, \frac{x}{A_1\sigma_5^2})$
Then $\frac{dF}{dx}= m(u(x))^{m-1}\frac{du}{dx}$ $\frac{du}{dx}= \frac{exp(\delta)}{\sqrt{\alpha^2\sigma_2^2\sigma_4^2}} \frac{dv}{dx}$ $\frac{dv}{dx}= \sum_{d= 0}^\infty W_{-d-0.5,0}(2\delta)\cdot \frac{d\gamma(1+ d, \frac{x}{A_1\sigma_5^2})}{dx}\frac{1}{A_1\sigma_5^2}$
So $\frac{df}{dx}= m\left(\frac{exp(\delta)}{\sqrt{\alpha^2\sigma_2^2\sigma_4^2}}\sum_{d= 0}^\infty W_{-d-0.5,0}(2\delta)\cdot \gamma(1+ d, \frac{x}{A_1\sigma_5^2})\right)\frac{exp(\delta)}{\sqrt{\alpha^2\sigma_2^2\sigma_4^2}}\sum_{d= 0}^\infty W_{-d-0.5,0}(2\delta)\cdot \frac{d\gamma(1+ d, \frac{x}{A_1\sigma_5^2})}{dx}\frac{1}{A_1\sigma_5^2}$