Question

Find the derivative of y=3x/(x^2 + 1)

Answer 1

To find the derivative of

$y=\frac{3x}{x^2+1},$

we will use the division rule for derivatives.

The division rule states that:

$(\frac{f(x)}{g(x)})^{\prime}=\frac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{g(x)^2}.$

Therefore, the derivative of the given quotient is:

$y^{\prime}=\frac{(3x)^{\prime}(x^2+1)-(x^2+1)^{\prime}(3x)}{(x^2+1)^2}.$

Simplifying the above result we get:

$\begin{gathered} y^{\prime}=\frac{3(x^2+1)-(2x\cdot3x)}{(x^2+1)^2}=\frac{3(x^2+1)-6x^2}{(x^2+1)^2}=\frac{3(x^2+1-2x^2)}{(x^2+1)^2} \\ =\frac{3(1-x^2)}{(x^2+1)^2}. \end{gathered}$

Answer:

$y^{\prime}=\frac{3(1-x^2)}{(x^2+1)^2}.$