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Find dy
- Thread starter kggirl
- Start date
Hello, kggirl!
Sheesh . . . neither of you ever heard of the Quotient Rule ??
Therefore: .\(\displaystyle \L dy\;=\;\frac{2x}{(1\,+\,x^2)^2}\,dx\)
Sheesh . . . neither of you ever heard of the Quotient Rule ??
Differentiate: .\(\displaystyle \L\frac{dy}{dx}\:=\:\frac{(1\,+\,x^2)\cdot2x\,-\,x^2\cdot2x}{(1\,+\,x^2)^2}\:=\:\frac{2x\,+\,2x^3\,-\,2x^3}{(1\,+\,x^2)^2}\:=\:\frac{2x}{(1\,+\,x^2)^2}\)If \(\displaystyle y\:=\:\frac{x^2}{1\,+\,x^2}\), find \(\displaystyle dy\).
Therefore: .\(\displaystyle \L dy\;=\;\frac{2x}{(1\,+\,x^2)^2}\,dx\)