mpute $$ \frac{d}{d x}\left(\left(x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+x^{\pi+13}+\sqrt[5]{x^{4}}\right)\left(x^{1+\sqrt{3}}+50 x^{2}+x-900\right)\right) $$. No need to s.

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$$
\begin{aligned}
f(x) &= x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + x^{\pi+13} + \sqrt[5]{x^4} = x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + x^{\pi+13} + x^{\frac{4}{5}},\[1mm]
g(x) &= x^{1+\sqrt{3}} + 50x^2 + x - 900.
\end{aligned}
$$

Differentiate $$f(x)$$ term by term:
$$
\begin{aligned}
f'(x) &= 7x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 + (\pi+13)x^{\pi+12} + \frac{4}{5}x^{\frac{4}{5}-1} \
&= 7x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 + (\pi+13)x^{\pi+12} + \frac{4}{5}x^{-\frac{1}{5}}.
\end{aligned}
$$

Differentiate $$g(x)$$ term by term:
$$
\begin{aligned}
g'(x) &= (1+\sqrt{3})x^{(1+\sqrt{3})-1} + 2\cdot50x + 1 \
&= (1+\sqrt{3})x^{\sqrt{3}} + 100x + 1.
\end{aligned}
$$

Using the product rule, we have:
$$
\frac{d}{dx}\Big(f(x)g(x)\Big) = f'(x)g(x) + f(x)g'(x).
$$

Substitute back:
$$
\begin{aligned}
\frac{d}{dx}\Big(&f(x)g(x)\Big) = \
&\Bigl(7x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 + (\pi+13)x^{\pi+12} + \frac{4}{5}x^{-\frac{1}{5}}\Bigr) \
&\quad \times \Bigl(x^{1+\sqrt{3}}+50x^2+x-900\Bigr)\[1mm]
&+\Bigl(x^7+x^6+x^5+x^4+x^3+x^2+x+x^{\pi+13}+x^{\frac{4}{5}}\Bigr)\
&\quad \times \Bigl((1+\sqrt{3})x^{\sqrt{3}}+100x+1\Bigr).
\end{aligned}
$$

This is the derivative of the given function.
The derivative of the function is: $$ \begin{aligned} \frac{d}{dx}\Big(&f(x)g(x)\Big) = \ &\Bigl(7x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 + (\pi+13)x^{\pi+12} + \frac{4}{5}x^{-\frac{1}{5}}\Bigr) \ &\quad \times \Bigl(x^{1+\sqrt{3}}+50x^2+x-900\Bigr)\[1mm] &+\Bigl(x^7+x^6+x^5+x^4+x^3+x^2+x+x^{\pi+13}+x^{\frac{4}{5}}\Bigr)\ &\quad \times \Bigl((1+\sqrt{3})x^{\sqrt{3}}+100x+1\Bigr). \end{aligned} $$

mpute . No need to s.

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