Question

find dy.

( y=\frac{3 x}{1+x^{2}} )
( d y=square d x )

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if \(y = \frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here, \(u = 3x\), so \(u^\prime=3\); \(v = 1 + x^{2}\), so \(v^\prime = 2x\).

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Step2: Simplify the numerator

Expand the numerator: \(3(1 + x^{2})-3x(2x)=3 + 3x^{2}-6x^{2}=3 - 3x^{2}=3(1 - x^{2})\)
So, \(y^\prime=\frac{3(1 - x^{2})}{(1 + x^{2})^{2}}\)
Since \(dy=y^\prime dx\)

Answer:

\(\frac{3(1 - x^{2})}{(1 + x^{2})^{2}}\)