find dy.\ny = \\frac{2x}{1 + 5x^{2}}\n\ndy = \\square dx
find dy.\ny = \\frac{2x}{1 + 5x^{2}}\n\ndy = \\square dx
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 = 2x), so (u^\prime=2); (v = 1 + 5x^{2}), so (v^\prime = 10x).
Step2: Substitute into the quotient rule formula
[ \begin{align*} y^\prime&=\frac{(2)(1 + 5x^{2})-(2x)(10x)}{(1 + 5x^{2})^{2}}\ &=\frac{2+10x^{2}-20x^{2}}{(1 + 5x^{2})^{2}}\ &=\frac{2-10x^{2}}{(1 + 5x^{2})^{2}} \end{align*} ] Since (dy=y^\prime dx), then (dy=\frac{2 - 10x^{2}}{(1 + 5x^{2})^{2}}dx)
Answer:
(\frac{2-10x^{2}}{(1 + 5x^{2})^{2}})