Question

Calcule a derivada imlicita das equacoes √x^2+y^2 = tan (x)

Answer 1

Olá

$\displaystyle \mathsf{ \sqrt{x^2+y^2}=tg(x) }\\\\\\\text{Derivada de raiz quadrada:~} \mathsf{\frac{u'}{2 \sqrt{u} } }\\\\\\\text{Derivando implicitamente}\\\\\\\mathsf{ \frac{2x+2yy'}{2 \sqrt{x^2+y^2} }=sec^2(x) }\\\\\\\text{Colocando o 2 em evidencia no numerador}\\\\\\\mathsf{ \frac{2(x+yy')}{2 \sqrt{x^2+y^2} }=sec^2(x)}\\\\\\\text{Simplifica}\\\\\\\mathsf{ \frac{\diagup\!\!\!\!2(x+yy')}{\diagup\!\!\!\!\!2 \sqrt{x^2+y^2} }=sec^2(x)}\\\\\\\mathsf{ \frac{x+yy'}{ \sqrt{x^2+y^2} }=sec^2(x)}$


Passa o denominador para o outro lado multiplicando

$\displaystyle \mathsf{ x+yy' =sec^2(x)\cdot{ \sqrt{x^2+y^2}}}}\\\\\\\mathsf{ yy' =sec^2(x)\cdot{ \sqrt{x^2+y^2}~-~x}}}\\\\\\\\\text{Passa o 'y' para o outro lado dividindo}\\\\\\\boxed{\mathsf{ y' = \frac{sec^2(x)\cdot{ \sqrt{x^2+y^2}~-~x}}{y} }}}$