# Help: derivatives 2

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

2 years, 3 months ago

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$$xy = 1$$

$$ydx + xdy = 0$$

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

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

$$\frac{dx}{\sqrt{x^4+ 1}} + \frac{dy}{\sqrt{y^4 + 1}} = 0$$

- 2 years, 3 months ago

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.

- 2 years, 3 months ago

OK......

- 2 years, 3 months ago

I don't know how to thank you always attending my doubts as soon as i post them. I am very grateful to you.

- 2 years, 3 months ago

No problem dude ... :-)

- 2 years, 3 months ago

Comment deleted Mar 10, 2016

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.

- 2 years, 3 months ago

Comment deleted Mar 11, 2016

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.

- 2 years, 3 months ago

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