Derive a função
a função g(x)=\sqrt{\frac{\cos (\sqrt{x})}{1+\operatorname{sen}(\sqrt{x})}}

Passo 1

O enunciado pede a derivada de
g(x)=\sqrt{\frac{\cos\bigl(\sqrt{x}\bigr)}{1+\sin\bigl(\sqrt{x}\bigr)}}.
Para resolver, vamos usar:
  1. A regra da cadeia aplicada à raiz quadrada,
  2. A regra do quociente para o termo dentro da raiz,
  3. Derivadas de \cos, \sin e de \sqrt{x}.

Passo 2

Defina
h(x)=\frac{\cos\bigl(\sqrt{x}\bigr)}{1+\sin\bigl(\sqrt{x}\bigr)},
de modo que
g(x)=\sqrt{h(x)}= \bigl(h(x)\bigr)^{1/2}.
Pela regra da cadeia,
g'(x)=\frac{1}{2}\bigl(h(x)\bigr)^{-1/2}\,h'(x) =\frac{h'(x)}{2\sqrt{h(x)}}.

Passo 3

Para achar h'(x), note que h é quociente de
\text{num}(x)=\cos\bigl(\sqrt{x}\bigr), \quad \text{den}(x)=1+\sin\bigl(\sqrt{x}\bigr).
Seja ainda
u(x)=\sqrt{x}, \quad u'(x)=\frac{1}{2\sqrt{x}}.
Então
\text{num}'(x)=-\sin\bigl(u(x)\bigr)\,u'(x), \quad \text{den}'(x)=\cos\bigl(u(x)\bigr)\,u'(x).
Pela regra do quociente,
h'(x) =\frac{\text{num}'(x)\,\text{den}(x)-\text{num}(x)\,\text{den}'(x)} {\bigl(\text{den}(x)\bigr)^2}.

Passo 4

Substituindo e fatorando u'(x):
h'(x) =\frac{u'(x)\Bigl[-\sin(u)\,(1+\sin(u))-\cos^2(u)\Bigr]}{(1+\sin(u))^2},
onde escrevemos u=\sqrt{x}. Note que
-\sin(u)\,(1+\sin(u))-\cos^2(u) =-\sin(u)-\sin^2(u)-\cos^2(u) =-\sin(u)-1 =-(1+\sin(u)).
Logo
h'(x) =\frac{u'(x)\bigl[-(1+\sin(u))\bigr]}{(1+\sin(u))^2} =-\frac{u'(x)}{1+\sin(u)} =-\frac{1}{2\sqrt{x}\,\bigl(1+\sin(\sqrt{x})\bigr)}.

Passo 5

Substituindo h'(x) em
g'(x)=\frac{h'(x)}{2\sqrt{h(x)}}
obtemos
g'(x) =-\frac{1}{2\sqrt{h(x)}}\;\frac{1}{2\sqrt{x}\,(1+\sin(\sqrt{x}))} =-\frac{1}{4\sqrt{x}\,(1+\sin(\sqrt{x}))\,\sqrt{h(x)}}.
Mas \sqrt{h(x)}=g(x) ou ainda
\sqrt{h(x)} =\sqrt{\frac{\cos(\sqrt{x})}{1+\sin(\sqrt{x})}}.
Para deixar tudo em termos de x, podemos combinar fatores:
(1+\sin(\sqrt{x}))\,\sqrt{h(x}) =(1+\sin(\sqrt{x}))\, \frac{\sqrt{\cos(\sqrt{x})}}{\sqrt{1+\sin(\sqrt{x})}} =\sqrt{\cos(\sqrt{x})}\,\sqrt{1+\sin(\sqrt{x})}.
Assim
g'(x) =-\frac{1}{4\,\sqrt{x}\,\sqrt{\cos(\sqrt{x})}\,\sqrt{1+\sin(\sqrt{x})}}.

Resposta

g'(x) =-\frac{1}{4\,\sqrt{x}\,\sqrt{\cos\bigl(\sqrt{x}\bigr)}\, \sqrt{1+\sin\bigl(\sqrt{x}\bigr)}}.

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