$\text { Find the derivative of the function } f ( x ) = \ln \left( \frac { 3 x } { x ^ { 2 } + 4 } \right) \text {. }$
A) $f ^ { t } ( x ) = \frac { 1 } { x } + \frac { 2 x } { x ^ { 2 } + 4 }$
B)$f ^ { t} ( x ) = \frac { 4 - x ^ { 2 } } { x \left( x ^ { 2 } + 4 \right) }$
C)$f ^ { t} ( x ) = \frac { 1 } { x } - \frac { 2 x } { x ^ { 2 } + 4 }$
D) $f ^ { t } ( x ) = \frac { 3 } { x } + \frac { 2 x } { x ^ { 2 } + 4 }$
E) $f ^ { t } ( x ) = \frac { 1 } { 3 x } - \frac { 2 x } { x ^ { 2 } + 4 }$

Question

$\text { Find the derivative of the function } f ( x ) = \ln \left( \frac { 3 x } { x ^ { 2 } + 4 } \right) \text {. }$ 
A) $f ^ { t } ( x ) = \frac { 1 } { x } + \frac { 2 x } { x ^ { 2 } + 4 }$
B)$f ^ { t} ( x ) = \frac { 4 - x ^ { 2 } } { x \left( x ^ { 2 } + 4 \right) }$
C)$f ^ { t} ( x ) = \frac { 1 } { x } - \frac { 2 x } { x ^ { 2 } + 4 }$
D) $f ^ { t } ( x ) = \frac { 3 } { x } + \frac { 2 x } { x ^ { 2 } + 4 }$
E) $f ^ { t } ( x ) = \frac { 1 } { 3 x } - \frac { 2 x } { x ^ { 2 } + 4 }$

Quizplus · Accepted Answer

This problem requires the use of the quotient rule and chain rule. Using the quotient rule, we have:

$$f'(x) = \frac{\frac{3}{x(x^2+4)}(x^2+4)-\frac{6x^2}{(x^2+4)^2}(3x)}{\left(\frac{3x}{x^2+4}\right)^2}$$

Simplifying and factoring out, we get:

$$f'(x) = \frac{\frac{1}{x}\left(\frac{4-x^2}{x^2+4}\right)}{\frac{9x^2}{(x^2+4)^2}}$$

Simplifying further, we get:

$$f'(x) = \frac{4-x^2}{x(x^2+4)} \cdot \frac{(x^2+4)^2}{9x^2}$$

$$f'(x) = \frac{4-x^2}{9x(x^2+4)}$$

Therefore, the correct choice is C.

Find the derivative of the function f(x)=ln⁡(3xx2+4). \text { Find the derivative of the function } f ( x ) = \ln \left( \frac { 3 x } { x ^ { 2 } + 4 } \right) \text {. } Find the derivative of the function f(x)=ln(x2+43x​).

$\text { Find the derivative of the function } f ( x ) = \ln \left( \frac { 3 x } { x ^ { 2 } + 4 } \right) \text {. }$