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Help: derivatives 2

If \( y = \dfrac1x \), prove that \( \dfrac{dy}{\sqrt{y^4+1}} + \dfrac{dx}{\sqrt{x^4+1}} = 0 \).

Note by Akhilesh Prasad
6 months, 3 weeks ago

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\( xy = 1\)

\( ydx + xdy = 0 \)

\( \frac{ydx}{\sqrt{x^2 + y^2}} + \frac{xdy}{\sqrt{x^2 + y^2}} = 0 \)

\( \frac{dx}{\sqrt{\frac{x^2}{y^2} + 1}} + \frac{dy}{\sqrt{\frac{y^2}{x^2} + 1}} = 0 \)

\( \frac{dx}{\sqrt{x^4+ 1}} + \frac{dy}{\sqrt{y^4 + 1}} = 0 \) Siddhartha Srivastava · 6 months, 3 weeks ago

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@Siddhartha Srivastava Thanks man, that was very helpful.@Rishabh Cool, this is what I was looking for, what i wanted was that you proceed purely from the given expression and arrive at the result that you are required to prove and not use the result required and substitute to it a condition derived from the given expression. Akhilesh Prasad · 6 months, 3 weeks ago

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@Akhilesh Prasad OK...... Rishabh Cool · 6 months, 3 weeks ago

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@Rishabh Cool I don't know how to thank you always attending my doubts as soon as i post them. I am very grateful to you. Akhilesh Prasad · 6 months, 3 weeks ago

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@Akhilesh Prasad No problem dude ... :-) Rishabh Cool · 6 months, 3 weeks ago

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@Rishabh Cool Shouldn't you be proving it from the relation and not by putting the value of \(dy\) & \(y\), cause wouldn't that just mean that you are using the result to prove the result. Akhilesh Prasad · 6 months, 3 weeks ago

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@Rishabh Cool I was not able to find the words for what i wanted to say lemme try it again. What i am saying is shouldn't you solely use the given expression to prove \( \dfrac{dy}{\sqrt{y^4+1}} + \dfrac{dx}{\sqrt{x^4+1}} = 0 \) and not substitute to it what the expression leads to, to get the proof. Akhilesh Prasad · 6 months, 3 weeks ago

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