Find the derivative of the function.
Select the correct answer.
$$f ( x ) = \frac { 2 \sqrt { x } } { x ^ { 2 } + 9 }$$
A)
$\frac { - 3 x ^ { 2 } + 9 } { \sqrt { x } \left( x ^ { 2 } + 9 \right) ^ { 2 } }$
B)
$\frac { 1 } { 2 x \sqrt { x } \left( x ^ { 2 } + 9 \right) }$
C)
$\frac { 1 } { 2 x \sqrt { x } }$
D)
$\frac { - 3 x ^ { 2 } + 9 } { \sqrt { x } \left( x ^ { 2 } + 9 \right) }$

Question

Quizplus · Accepted Answer

The derivative of the function can be found using the quotient rule, which is:$$f'(x) = \frac{(x^2 + 9)\frac{d}{dx}(2\sqrt{x}) - 2\sqrt{x}\frac{d}{dx}(x^2 + 9)}{(x^2 + 9)^2}$$Simplifying this expression, we get:$$f'(x) = \frac{(x^2 + 9)\frac{1}{\sqrt{x}} - 2\sqrt{x}(2x)}{(x^2 + 9)^2}$$$$f'(x) = \frac{\frac{x^2 + 9}{\sqrt{x}} - 4x\sqrt{x}}{(x^2 + 9)^2}$$$$f'(x) = \frac{x^2 + 9 - 4x^2}{(x^2 + 9)^2\sqrt{x}}$$$$f'(x) = \frac{-3x^2 + 9}{(x^2 + 9)^2\sqrt{x}}$$

Find the Derivative of the Function f(x)=2xx2+9f ( x ) = \frac { 2 \sqrt { x } } { x ^ { 2 } + 9 }f(x)=x2+92x​​

Find the Derivative of the Function $f ( x ) = \frac { 2 \sqrt { x } } { x ^ { 2 } + 9 }$