Question

Calcular a derivada dy/dx
y=raiz de u ; u=x^2+2x-3.
Conta detalhada por favor.
resposta dy/dx=x+1/raiz de x^2+2x-3

Answer 1

$f_{(x)}=y=\sqrt{u}\\\\ f_{(x)}=\sqrt{x^2+2x-3}\\\\ f_{(x)}=(x^2+2x-3)^{\frac{1}{2}}\\\\\\ Derivada:\\\\ \frac{dy}{dx}=f'_{(x)}\\\\\\ Derivando\ pela\ regra\ da\ cadeia\\\\ Derivada\ de\ fora\ vezes\ a\ derivada\ de\ dentro:\\\\ f'_{(x)}=((x^2+2x^1-3)^{\frac{1}{2}})'\times(x^2+2x^1-3)'}\\\\ f'_{(x)}=\frac{1}{2}(x^2+2x^1-3)^{(\frac{1}{2}-1)}\times(2x^{(2-1)}+2.1.x^{(1-1)}-0)}\\\\ f'_{(x)}=\frac{1}{2}(x^2+2x^1-3)^{-\frac{1}{2}}\times(2x^{1}+2.1.x^{0})}$

$f'_{(x)}=\frac{1}{2}\frac{1}{(x^2+2x^1-3)^{\frac{1}{2}}}\times(2x+2.1.1)}\\\\ f'_{(x)}=\frac{1}{2\sqrt{x^2+2x^1-3}}\times(2x+2)}\\\\ f'_{(x)}=\frac{(2x+2)}{2\sqrt{x^2+2x^1-3}}}\\\\ f'_{(x)}=\frac{2(x+1)}{2\sqrt{x^2+2x^1-3}}}\\\\ f'_{(x)}=\frac{\not2(x+1)}{\not2\sqrt{x^2+2x^1-3}}}\\\\ \boxed{f'_{(x)}=\frac{x+1}{\sqrt{x^2+2x^1-3}}}}$

Bons estudos!