A calculus problem by Divyansh Khatri

By applying Chain Rule again and again :

$\begin{aligned} & = 10(5)(1+(2-(6+7x^{4})^{9})^{3})^{4} \left[1+(2-(6+7x^{4})^{9})^{3} \right]' \\ & = 50 (1+(2-(6+7x^{4} )^{9})^{3})^{4} \left[ 0+3 (2-(6+7x^{4})^{9})^{2} + \left[ 2-(6+7x^{4})^{9} \right] ' \right] \\ & = 150 (1+(2-(6+7x^{4} )^{9})^{3})^{4} (2-(6+7x^{4})^{9})^{2} + \left[ 0-9(6+7x^4)^8 \left[ 6+7x^4\right] ' \right] \\ & = -1350 (1+(2-(6+7x^{4} )^{9})^{3})^{4} (2-(6+7x^{4})^{9})^{2} (6+7x^4)^8 28x^3 \\ & = -37800x^3 (1+(2-(6+7x^{4} )^{9})^{3})^{4} (2-(6+7x^{4})^{9})^{2} (6+7x^4)^8 \end{aligned}$

Now, as suggested, we can use Wolfram alpha to find last $9$ digits of the above expression at $x=1$ .

$\Rightarrow \boxed{818000000}$