To find the derivative of the function [tex]\( f(x) = \frac{6}{x+4} + \frac{3}{x^2} \)[/tex], we will use the rules of differentiation.
Firstly, let's consider the first term [tex]\( \frac{6}{x+4} \)[/tex].
Rewrite it as [tex]\( 6 \cdot (x+4)^{-1} \)[/tex].
Now, using the power rule and the chain rule:
[tex]\[ \frac{d}{dx}[6 \cdot (x+4)^{-1}] = 6 \cdot -1 \cdot (x+4)^{-2} \cdot \frac{d}{dx}[x+4] \][/tex]
[tex]\[ = 6 \cdot -1 \cdot (x+4)^{-2} \cdot 1 = -6 \cdot (x+4)^{-2} \][/tex]
[tex]\[ = \frac{-6}{(x+4)^2} \][/tex]
Next, consider the second term [tex]\( \frac{3}{x^2} \)[/tex].
Rewrite it as [tex]\( 3 \cdot x^{-2} \)[/tex].
Again, using the power rule:
[tex]\[ \frac{d}{dx}[3 \cdot x^{-2}] = 3 \cdot -2 \cdot x^{-3} \][/tex]
[tex]\[ = -6 \cdot x^{-3} \][/tex]
[tex]\[ = \frac{-6}{x^3} \][/tex]
Now, sum the derivatives of the two terms:
[tex]\[ \frac{d}{dx} \left( \frac{6}{x+4} + \frac{3}{x^2} \right) = \frac{-6}{(x+4)^2} + \frac{-6}{x^3} \][/tex]
Combining these results, we get the derivative:
[tex]\[ f'(x) = \frac{-6}{(x+4)^2} + \frac{-6}{x^3} \][/tex]
Comparing with the given options:
- [tex]\(\frac{6}{x^2+8 x+16}+\frac{6}{x}\)[/tex]
- [tex]\(6+\frac{3}{2 x}\)[/tex]
- [tex]\(\frac{-6}{(x+4)^2}+\frac{3}{x^3}\)[/tex]
- [tex]\(\frac{-6}{x^2+8 x+16}-\frac{6}{x^3}\)[/tex]
- None of these
The correct derivative [tex]\( f'(x) = \frac{-6}{(x+4)^2} + \frac{-6}{x^3} \)[/tex] does not match any of the provided options exactly. However, the option [tex]\(\frac{-6}{(x+4)^2}+\frac{3}{x^3}\)[/tex] is similar but incorrect.
Therefore, the correct answer is:
[tex]\[ \boxed{\text{None of these}} \][/tex]
