Find $d y$
\[
y=\frac{7 x}{1+3 x^{2}}
\]

Answer

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Answer

The derivative of the function $y=\frac{7 x}{1+3 x^{2}}$ is $\boxed{\frac{7 - 21x^2}{(1+3x^2)^2}}$.

Steps

Step 1 ：Define the functions $f(x) = 7x$ and $g(x) = 1 + 3x^2$.

Step 2 ：Calculate the derivatives of these functions, $f'(x) = 7$ and $g'(x) = 6x$.

Step 3 ：Use the quotient rule to find the derivative of $y$, which is $dy = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$.

Step 4 ：Substitute the functions and their derivatives into the formula to get $dy = \frac{7(1 + 3x^2) - 7x(6x)}{(1 + 3x^2)^2}$.

Step 5 ：Simplify the expression to get the final answer: $dy = \frac{7 - 21x^2}{(1+3x^2)^2}$.

Step 6 ：The derivative of the function $y=\frac{7 x}{1+3 x^{2}}$ is $\boxed{\frac{7 - 21x^2}{(1+3x^2)^2}}$.