Question

Derivar y = r/ (r² + 1)^1/2

Answer 1

$y=\dfrac{r}{(r^{2}+1)^{1/2}}$


Aplicando a Regra do Quociente e a Regra da Cadeia, temos

$\dfrac{dy}{dr}=\dfrac{d}{dr}\left[\dfrac{r}{(r^{2}+1)^{1/2}} \right ]\\ \\ \\ \dfrac{dy}{dr}=\dfrac{\frac{d}{dr}(r)\cdot (r^{2}+1)^{1/2}-r\cdot \frac{d}{dr}[(r^{2}+1)^{1/2}]}{[(r^{2}+1)^{1/2}]^{2}}\\ \\ \\ \dfrac{dy}{dr}=\dfrac{1\cdot (r^{2}+1)^{1/2}-r\cdot \frac{1}{2}(r^{2}+1)^{(1/2)-1}\cdot \frac{d}{dr}(r^{2}+1)}{[(r^{2}+1)^{1/2}]^{2}}\\ \\ \\ \dfrac{dy}{dr}=\dfrac{(r^{2}+1)^{1/2}-r\cdot \frac{1}{2}(r^{2}+1)^{-1/2}\cdot (2r)}{(r^{2}+1)^{(1/2)\,\cdot\, 2}}\\ \\ \\ \dfrac{dy}{dr}=\dfrac{(r^{2}+1)^{1/2}-r^{2}\,(r^{2}+1)^{-1/2}}{(r^{2}+1)}$

 
Multiplicando o numerador e o denominador por $(r^{2}+1)^{1/2},$ temos

$\dfrac{dy}{dr}=\dfrac{[(r^{2}+1)^{1/2}-r^{2}\,(r^{2}+1)^{-1/2}]\cdot (r^{2}+1)^{1/2}}{(r^{2}+1)\cdot (r^{2}+1)^{1/2}}\\ \\ \\ \dfrac{dy}{dr}=\dfrac{(r^{2}+1)^{(1/2)+(1/2)}-r^{2}\,(r^{2}+1)^{-(1/2)+(1/2)}}{(r^{2}+1)^{1+(1/2)}}\\ \\ \\ \dfrac{dy}{dr}=\dfrac{(r^{2}+1)-r^{2}}{(r^{2}+1)^{3/2}}\\ \\ \\ \dfrac{dy}{dr}=\dfrac{1}{(r^{2}+1)^{3/2}}$