[tex]\cos{(x+y)}=1-x^2\ \therefore\ x+y=\arccos{(1-x^2)}\ \therefore\\\\ \boxed{y(x)=\arccos{(1-x^2)}-x}[/tex]
A derivada é separável. Além disso, devemos lembrar da Regra da Cadeia na derivada de arccos(1 - x²):
[tex]\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\arccos{(1-x^2)}\bigg)\bigg[\dfrac{d}{dx}(1-x^2)\bigg]-\dfrac{d}{dx}(x)\ \therefore[/tex]
A derivada de arccos(u) é tabelada, e vale:
[tex]\boxed{\dfrac{d}{du}\big(\arccos{u}\big)=-\dfrac{1}{\sqrt{1-u^2}}}[/tex]
Dessa forma:
[tex]\dfrac{dy}{dx}=\bigg(-\dfrac{1}{\sqrt{1-(1-x^2)^2}}\bigg)\bigg[-2x\bigg]-1\ \therefore[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-(1+x^4-2x^2)}}-1=\dfrac{2x}{\sqrt{2x^2-x^4}}-1\ \therefore[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{2x}{x\sqrt{2-x^2}}-1\ \therefore\ \boxed{\dfrac{dy}{dx}=\dfrac{2}{\sqrt{2-x^2}}-1}[/tex]
