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yesteryday was my class test and I found this question.

Find $\displaystyle \frac{dy}{dx}$ when, $\displaystyle y \arcsin x - x \arctan y = 1$

I have read the question for arctanx as $1/1 + x^2$. But that is applied only, when the question is

Find difference is along x (dy/dx) and the function has the x as the variable.

How would I solve the above equation?

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up vote 2 down vote accepted

Using Chain Rule we have , $$\frac{dy}{dx}\cdot\arcsin x+y\cdot\frac{d(\arcsin x)}{dx}-\frac{dx}{dx}\cdot \arctan y -x\cdot\frac{d(\arctan y)}{dy}\frac{dy}{dx}=\frac{d(1)}{dx}$$

$$\frac{dy}{dx}\cdot\arcsin x+y\cdot\frac1{\sqrt{1-x^2}}-\arctan y -x\cdot\frac1{1+y^2}\frac{dy}{dx}=0$$

Now group the terms containing $\displaystyle \frac{dy}{dx}$

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I can sense a product rule in it! :) Simple and easy, but can you please elaborate how can I use d(arctany)/dy while I need to be using d/dx? I mean how was chain rule applied which values changed to variables? – Afzaal Ahmad Zeeshan Dec 8 '13 at 9:42
@AfzaalAhmadZeeshan, $$\frac{d f(y)}{dx}=\frac{ d f(y)}{dy}\cdot\frac{dy}{dx}=f'(y)\cdot\frac{dy}{dx}$$ – lab bhattacharjee Dec 8 '13 at 9:44
Oh haha :D that was simple! Thanks alot sir g :) will accept it as answer in one minute! :) – Afzaal Ahmad Zeeshan Dec 8 '13 at 9:45

Use the total differentiation of an implicit function $F =\operatorname F(x,y) = 0$. You end with
$\displaystyle \frac {\operatorname dy}{\operatorname dx} = - \dfrac{ \frac {\operatorname dF}{\operatorname dx} }{ \frac { \operatorname dF }{\operatorname dy}}$
and you wil arrive to what lab bhattarcharjee gives.

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nice one! :) thanks for info..:) – Afzaal Ahmad Zeeshan Dec 8 '13 at 9:47
@AfzaalAhmadZeeshan. Glad to have helped – Claude Leibovici Dec 8 '13 at 9:49
@Arjang. Thanks for editing. Since I am almost blind, everything beside ASCII is difficult to me. – Claude Leibovici Dec 8 '13 at 9:56
@ClaudeLeibovici : it is my pleasure – Arjang Dec 8 '13 at 10:18

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