1. ## Calculus, dervatives

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.

2. Originally Posted by Dranalion
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 $x^2$ is, of course, 2x. By the quotient rule, the derivative of f(y)/g(y) is $(f'g- fg')/g$. Further, the derivative of arctan(y) is $\frac{1}{1+ y^2}$ so the derivative of $\frac{arctan(y)}{1+ y^2}$, with respect to y, is $\frac{\frac{1}{1+ y^2}(1+y^2)- arctan(y)(2y)}{(1+ y^2)^2}= \frac{1- y arctan(y)}{(1+ y^2)^2}$.
The derivative with respect to x, rather than y, is that times $\frac{dy}{dx}$ so we have $2x= \frac{1- y arctan(y)}{(1+ y^2)^2}\frac{dy}{dx}$.

3. I got this however, what have I done?

(1+y^2)*x^2 = arctan(y)
2x(1+y^2) + x^2(1+2yy') = y'/(1+y^2)
2x+2xy^2+x^2+2yy'x^2 = y'/(1+y^2)
2x+2xy^2+x^2 = y'/(1+y^2) - 2yy'x^2 = y'(1/(1+y^2) - 2yx^2)
y' = (2x+2xy^2+x^2) / (1/(1+y^2) - 2yx^2)

4. Hello, Dranalion!

Find $\frac{dy}{dx}$ for: . $x^2 \:=\: \frac{\arctan(y)}{1+y^2}$
First, I'd get rid of the quotient . . . $x^2(1+y^2) \:=\:\arctan(y)
$

And might as well get rid of the product . . . $x^2 + x^2y^2 \:=\:\arctan(y)$

Differentiate: . $2x + x^2\!\cdot\!2y\!\cdot\!y' + 2xy^2 \:=\:\frac{y'}{1+y^2} \quad\Rightarrow\quad 2x^2y\!\cdot\!y' + 2x(1 + y^2) \:=\:\frac{y'}{1+y^2}$

. . $2x^2y\!\cdot\!y'(1+y^2) + 2x(1+y^2)^2 \:=\:y' \quad\Rightarrow\quad y' - 2x^2y(1+y^2)\!\cdot\!y' \:=\:2x(1+y^2)^2$

Factor: . $\left[1 - 2x^2y(1+y^2)\right]\!\cdot\!y' \:=\:2x(1+y^2)^2 \quad\Rightarrow\quad y' \;=\;\frac{2x(1+y^2)^2}{1-2x^2y(1+y^2)}$

5. I got $\frac{-(1+y^2)}{2x}$