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find dy. y = \\frac{2x}{1 + 5x^{2}} dy = \\square dx

Question

find dy.
y = \frac{2x}{1 + 5x^{2}}

dy = \square dx

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

$$ LATEXBLOCK0 $$

Since \(dy=y^\prime dx\), then \(dy=\frac{2 - 10x^{2}}{(1 + 5x^{2})^{2}}dx\)

Answer:

\(\frac{2-10x^{2}}{(1 + 5x^{2})^{2}}\)