$\sin ^{-1}\left(\frac1{ x^{2}+y^{2}}\right) +\csc^{-1}\left( x^{2}+y^{2}\right) =\dfrac {\pi } {3}$ A(1,-1) NOKTASINDA $\frac{dx}{dy}$ yi bulunuz.

Question

\sin ^{-1}\left( x^{2}+y^{2}\right) ^{-1}+csc^{-1}\left( x^{2}+y^{2}\right) =\dfrac {\pi } {3}     A(1,-1) NOKTASINDA dx/dy  yi bulunuz.

Comments

Bu soru için bu kadar ısrar ettiğinize değecek mi acaba? Çok merak ediyorum. :)

---

DeğMeSeydi ugrasmazdiM :D

Answer 1

$d\left[sin^{-1}\left(\frac{1}{x^2+y^2}\right)\right]+d\left[csc^{-1}\left(x^2+y^2\right)\right]=d(\frac{\pi}{3})$

$\left(\frac{1}{x^2+y^2}\right)'\frac{1}{\sqrt{1-(x^2+y^2)^2}}dx+\left(\frac{1}{x^2+y^2}\right)'\frac{1}{\sqrt{1-(x^2+y^2)^2}}dy+\frac{(x^2+y^2)'}{(x^2+y^2)\sqrt{(x^2+y^2)^2-1}}dx+\frac{(x^2+y^2)'}{(x^2+y^2)\sqrt{(x^2+y^2)^2-1}}dy$=0

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

$\frac{d}{dx}\left[sin^{-1}\left(\frac{1}{x^2+y^2}\right)+csc^{-1}\left(x^2+y^2\right)\right]=$

$\frac{-\left(\frac{2x}{(x^2+y^2)^2}\right) \frac{1}{\sqrt{1-(x^2+y^2)^2}}-\frac{2x}{(x^2+y^2)\sqrt{(x^2+y^2)^2-1}}}{\left(\frac{2y}{(x^2+y^2)^2}\right) \frac{1}{\sqrt{1-(x^2+y^2)^2}}+\frac{2y}{(x^2+y^2)\sqrt{(x^2+y^2)^2-1}}}=$

$\frac{-\left(\frac{x}{x^2+y^2}\right) \frac{1}{\sqrt{1-(x^2+y^2)^2}}-\frac{x}{\sqrt{(x^2+y^2)^2-1}}}{\left(\frac{y}{x^2+y^2}\right) \frac{1}{\sqrt{1-(x^2+y^2)^2}}+\frac{y}{\sqrt{(x^2+y^2)^2-1}}}$

$(x^2+y^2)^2<1$ olmalıdır.