Hi, I am finding it difficult to undertake this question, can anyone please assist me?
Find dy/dx in terms of x and y for:
x^2 = arctan(y) / 1+y^2
Any help much appreciated,
Dranalion.
The derviative of $\displaystyle x^2$ is, of course, 2x. By the quotient rule, the derivative of f(y)/g(y) is $\displaystyle (f'g- fg')/g$. Further, the derivative of arctan(y) is $\displaystyle \frac{1}{1+ y^2}$ so the derivative of $\displaystyle \frac{arctan(y)}{1+ y^2}$, with respect to y, is $\displaystyle \frac{\frac{1}{1+ y^2}(1+y^2)- arctan(y)(2y)}{(1+ y^2)^2}= \frac{1- y arctan(y)}{(1+ y^2)^2}$.
The derivative with respect to x, rather than y, is that times $\displaystyle \frac{dy}{dx}$ so we have $\displaystyle 2x= \frac{1- y arctan(y)}{(1+ y^2)^2}\frac{dy}{dx}$.
Hello, Dranalion!
First, I'd get rid of the quotient . . . $\displaystyle x^2(1+y^2) \:=\:\arctan(y)Find $\displaystyle \frac{dy}{dx}$ for: .$\displaystyle x^2 \:=\: \frac{\arctan(y)}{1+y^2}$
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And might as well get rid of the product . . . $\displaystyle x^2 + x^2y^2 \:=\:\arctan(y)$
Differentiate: .$\displaystyle 2x + x^2\!\cdot\!2y\!\cdot\!y' + 2xy^2 \:=\:\frac{y'}{1+y^2} \quad\Rightarrow\quad 2x^2y\!\cdot\!y' + 2x(1 + y^2) \:=\:\frac{y'}{1+y^2} $
. . $\displaystyle 2x^2y\!\cdot\!y'(1+y^2) + 2x(1+y^2)^2 \:=\:y' \quad\Rightarrow\quad y' - 2x^2y(1+y^2)\!\cdot\!y' \:=\:2x(1+y^2)^2$
Factor: .$\displaystyle \left[1 - 2x^2y(1+y^2)\right]\!\cdot\!y' \:=\:2x(1+y^2)^2 \quad\Rightarrow\quad y' \;=\;\frac{2x(1+y^2)^2}{1-2x^2y(1+y^2)} $