find dy.\n\n( y=\frac{3 x}{1+x^{2}} )\n( d y=square d x )
find dy.\n\n( y=\frac{3 x}{1+x^{2}} )\n( d y=square d x )
Answer
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). [ \begin{align*} y^\prime&=\frac{3(1 + x^{2})-3x(2x)}{(1 + x^{2})^{2}}\ \end{align*} ]
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}})