Thread: 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.

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 is, of course, 2x. By the quotient rule, the derivative of f(y)/g(y) is . Further, the derivative of arctan(y) is so the derivative of , with respect to y, is .
The derivative with respect to x, rather than y, is that times so we have .

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)

Hello, Dranalion!

Find for: .

First, I'd get rid of the quotient . . .

And might as well get rid of the product . . .


Differentiate: .

. .


Factor: .

I got