I tried to solve it by:
Taking the derivative of both sides using Chain Rule:
$1 = \sec(\frac{1}{y})\tan(\frac{1}{y}) \frac{1}{y'}$
Multiplying both sides by the derivative of $y'$ to isolate $y'$:
$y'= \sec(\frac{1}{y})\tan(\frac{1}{y})$.
I do not feel this is the correct answer, so if you could tell me what I did wrong and show me the proper steps to find the correct answer, I would greatly appreciate it.
Best Answer
Yes, you have to use the chain rule. But you have a small mistake:
$$\frac{d(1/y)}{dx} = -\frac{y'}{y^2}$$
And not $1/y'$.
Edit:
$$1 =- \sec(1/y) \tan(1/y) \frac{y'}{y^2}$$
Multiplying both sides by $-y^2$, and dividing by $\sec(1/y) \tan(1/y)$, we get:
$$y' = -\frac{y^2}{\sec(1/y) \tan(1/y)}$$