Find the derivative of the function. $$f ( x ) = \frac { 2 \sqrt { x } } { x ^ { 2 } + 9 }$$
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

To find the derivative of $f(x) = \frac{2\sqrt{x}}{x^2 + 9}$, we use the quotient rule: $\left( \frac{g}{h} \right)' = \frac{g'h - gh'}{h^2}$, where $g(x) = 2\sqrt{x}$ and $h(x) = x^2 + 9$. The derivative of $g(x)$ is $g'(x) = \frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$, and the derivative of $h(x)$ is $h'(x) = 2x$. Substituting these into the quotient rule formula gives $\frac{\frac{1}{\sqrt{x}}(x^2 + 9) - 2\sqrt{x}(2x)}{(x^2 + 9)^2}$, which simplifies to $\frac{-3x^2 + 9}{\sqrt{x}(x^2 + 9)^2}$.

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 $f ( x ) = \frac { 2 \sqrt { x } } { x ^ { 2 } + 9 }$