Usando a regra da cadeia, calcule f^{\prime}(x) onde f(x) é igual a:
f(x)=\ln \operatorname{tg}(x / 2)

Passo 1

O enunciado pede a derivada de
f(x)=\ln\tan\bigl(\tfrac{x}{2}\bigr).
O conteúdo principal é a regra da cadeia, usando que
\frac{d}{dx}\ln u=\frac{u'}{u}
e
\frac{d}{dx}\tan v=\operatorname{sec}^2 v\cdot v'.

Passo 2

Seja u(x)=\tan\bigl(\tfrac{x}{2}\bigr). Então
f'(x)=\frac{u'(x)}{u(x)} =\frac{\frac{d}{dx}\tan\bigl(\tfrac{x}{2}\bigr)}{\tan\bigl(\tfrac{x}{2}\bigr)}.

Passo 3

Para derivar \tan\bigl(\tfrac{x}{2}\bigr) aplicamos regra da cadeia em \tan(v) com v=x/2:
\frac{d}{dx}\tan\bigl(\tfrac{x}{2}\bigr) =\operatorname{sec}^2\bigl(\tfrac{x}{2}\bigr)\cdot\frac{d}{dx}\bigl(\tfrac{x}{2}\bigr) =\operatorname{sec}^2\bigl(\tfrac{x}{2}\bigr)\cdot\tfrac12.
Logo
f'(x)=\frac{\tfrac12\,\operatorname{sec}^2\bigl(\tfrac{x}{2}\bigr)}{\tan\bigl(\tfrac{x}{2}\bigr)}.

Passo 4

Simplificamos
\frac{\operatorname{sec}^2\bigl(\tfrac{x}{2}\bigr)}{\tan\bigl(\tfrac{x}{2}\bigr)} =\frac{\tfrac1{\cos^2(\tfrac{x}{2})}}{\tfrac{\sin(\tfrac{x}{2})}{\cos(\tfrac{x}{2})}} =\frac{1}{\sin(\tfrac{x}{2})\cos(\tfrac{x}{2})}.
Assim
f'(x)=\frac12\cdot\frac{1}{\sin(\tfrac{x}{2})\cos(\tfrac{x}{2})}.

Passo 5

Usando a identidade
2\sin\bigl(\tfrac{x}{2}\bigr)\cos\bigl(\tfrac{x}{2}\bigr)=\sin(x),
vem
f'(x)=\frac{1}{2\sin(\tfrac{x}{2})\cos(\tfrac{x}{2})} =\frac{1}{\sin(x)}.

Resposta

f'(x)=\csc(x).

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