Question

Determine a derivada da função f(x) raiz quadrada de sen (x) cos^3 (x)

Answer 1

Acompanhe a regra da cadeia e a do produto:

$\displaystyle y = \sqrt{\sin(x)\cdot \cos^3(x)} \\ \\ \\ y' = (\sin(x)\cdot \cos^3(x))^{1/2} \\ \\ \\ y' = \frac{1}{2} \cdot (\sin(x)\cdot \cos^3(x))^{1/2-1} \cdot (\sin(x)\cdot \cos^3(x))' \\ \\ \\ y' = \frac{1}{2} \cdot (\sin(x)\cdot \cos^3(x))^{-1/2} \cdot (\, \sin'(x)\cos^3(x)+\sin(x)\cos^3(x)' \,) \\ \\ \\ y' = \frac{1}{2} \cdot \frac{1}{(\sin(x)\cdot \cos^3(x))^{1/2}} \cdot (\, \sin'(x)\cos^3(x)+\sin(x)\cos^3(x)' \,)$

$\displaystyle y' = \frac{1}{2} \cdot \frac{1}{\sqrt{\sin(x)\cdot \cos^3(x)}} \cdot (\, \sin'(x)\cos^3(x)+\sin(x)\cos^3(x)' \,) \\ \\ \\ y' = \frac{1}{2\sqrt{\sin(x)\cdot \cos^3(x)}} \cdot (\, \sin'(x)\cos^3(x)+\sin(x)\cos^3(x)' \,)$


O termo sen'(x)·cos³(x) + sen(x)·cos³(x)':


$\sin'(x)\cos^3(x)+\sin(x)\cos^3(x)' \\ \\ \\ \cos(x)\cos^3(x)+\sin(x) \cdot 3 \cdot \cos^{3-1}(x)\cdot \cos(x)' \\ \\ \\ \cos(x)\cos^3(x)+\sin(x) \cdot 3 \cdot \cos^{2}(x)\cdot (-\sin(x)) \\ \\ \\ \cos(x)\cos^3(x)-\sin(x) \cdot 3 \cdot \cos^{2}(x)\cdot \sin(x) \\ \\ \\ \cos^4(x)-3\sin^2(x)\cos^2(x)$

Continuando:

$\displaystyle y' = \frac{1}{2\sqrt{\sin(x)\cdot \cos^3(x)}} \cdot (\, \sin'(x)\cos^3(x)+\sin(x)\cos^3(x)' \,) \\ \\ \\ y' = \frac{1}{2\sqrt{\sin(x)\cdot \cos^3(x)}} \cdot \cos^4(x)-3\sin^2(x)\cos^2(x) \\ \\ \\ \boxed{\boxed{y' = \frac{\cos^4(x)-3\sin^2(x)\cos^2(x)}{2\sqrt{\sin(x)\cdot \cos^3(x)}}}}$