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The question is- Find $\frac{dy}{dx}$, if $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$

So, they assumed $x$=$a\cos^3\theta$, $y$=$a\sin^3\theta$

And then, $\frac{dy}{dx}$=$\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$ = $-\tan\theta$

But, if I were to interchange the values of $x$ and $y$, then, the answer would come out to be $-\cot\theta$. So, why do different values of slope come? Is the second answer correct too?

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    $\begingroup$ The $\theta $ in one case (given in your question) and the $\theta$ in other case (suggested in question) are different. If we use different symbols $\theta, \theta'$ for them then one can easily see that $\tan \theta=\cot \theta'$ so that the derivative is same using any substitution. The confusion arises only by using the same parameter in both cases. $\endgroup$ – Paramanand Singh May 31 '17 at 6:45
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    $\begingroup$ I have taken the liberty to modify your title in order that this question can be accessed/retrieved more easily. $\endgroup$ – Jean Marie May 31 '17 at 7:01
  • $\begingroup$ @JeanMarie Yeah! That's really appreciated! $\endgroup$ – Abhishekstudent May 31 '17 at 7:43
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For $y=a\sin^3\theta,$ etc.

$$-\dfrac{\sin\theta}{\cos\theta}=-\dfrac{y^{1/3}}{x^{1/3}}$$

For $y=a\cos^3\theta,$ etc.

$$-\dfrac{\cos\theta}{\sin\theta}=-\dfrac{y^{1/3}}{x^{1/3}}$$

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