Let $$ f(x)=\left(7 x-8 x^{4}\right)(3+\sqrt{x}) $$. Find $$ f^{\prime}(x) $$

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We are given
$$
f(x) = \left(7x-8x^4\right)\left(3+\sqrt{x}\right).
$$
Define
$$
u(x)=7x-8x^4 \quad \text{and} \quad v(x)=3+\sqrt{x}.
$$
Then, using the product rule, we have
$$
f'(x)=u'(x)v(x)+u(x)v'(x).
$$

1. Differentiate $$ u(x)=7x-8x^4 $$:
   $$
   u'(x)=7-32x^3.
   $$

2. Differentiate $$ v(x)=3+\sqrt{x} $$. Recall that $$\sqrt{x}=x^{\frac{1}{2}}$$ and its derivative is
   $$
   \frac{d}{dx}x^{\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}.
   $$
   So,
   $$
   v'(x)=\frac{1}{2\sqrt{x}}.
   $$

Now, plug these into the product rule:
$$
f'(x) = \left(7-32x^3\right)\left(3+\sqrt{x}\right) + \left(7x-8x^4\right)\left(\frac{1}{2\sqrt{x}}\right).
$$

Thus, the derivative is
$$
\boxed{f'(x)=\left(7-32x^3\right)\left(3+\sqrt{x}\right) + \frac{7x-8x^4}{2\sqrt{x}}.}
$$
The derivative of $$ f(x) $$ is: $$ f'(x) = (7 - 32x^3)(3 + \sqrt{x}) + \frac{7x - 8x^4}{2\sqrt{x}} $$

Let . Find

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