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.

  • $\begingroup$ Note: $(1/y)' \neq 1/(y')$. Instead $(1/y)' = -y'/(y^2)$ $\endgroup$ – Graham Kemp Jun 17 '15 at 3:54
  • $\begingroup$ You implemented the chain rule incorrectly in the first step. It seems to me that you're thinking that the derivative of $\frac{1}{y(x)}$ is $\frac{1}{y'(x)}$, which is wrong. $\endgroup$ – MathNewbie Jun 17 '15 at 3:54
  • $\begingroup$ So if I change it to -y'/y^2 how would I move y' to the other side of the equation? $\endgroup$ – Sarah M Jun 17 '15 at 4:05

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 $1/y'$.


$$1 =- \sec(1/y) \tan(1/y) \frac{y'}{y^2}$$

Multiplying both sides by $-y^2$, and dividing by $\sec(1/y) \tan(1/y)$, we get:

$$y' = -\frac{y^2}{\sec(1/y) \tan(1/y)}$$

  • $\begingroup$ Can you show me how to proceed from there? $\endgroup$ – Sarah M Jun 17 '15 at 5:36
  • $\begingroup$ @SarahM: check the edit. $\endgroup$ – user230734 Jun 17 '15 at 9:44
  • $\begingroup$ To solve the problem you should express $y'$ as a function of $x$ $\endgroup$ – Marco Disce Apr 12 '16 at 18:41

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