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Exercícios-Derive
y=\operatorname{sen}(x)-\cos \left(\frac{2 x}{x+1}\right)+\operatorname{tg}\left(x^{2}\right)+\operatorname{cotg}(x)-\operatorname{sec} \left(\frac{x}{3}\right)+\operatorname{cossec}(2 \pi)
-\operatorname{arctg}(2 x)

Passo 1

O enunciado pede a derivada de
y = \operatorname{sen}(x)\;-\;\cos\!\Bigl(\tfrac{2x}{x+1}\Bigr)\;+\;\tan\bigl(x^{2}\bigr)\;+\;\operatorname{cotg}(x)\;-\;\operatorname{sec}\!\Bigl(\tfrac{x}{3}\Bigr)\;+\;\cossec(2\pi)\;-\;\operatorname{arctg}(2x).
O conteúdo principal envolve as derivadas das funções trigonométricas usuais (sen, cos, tan, cotg, sec, cossec), da função arctg e uso da regra da cadeia e da regra do quociente.

Passo 2

Derivada de \operatorname{sen}(x) e de -\cos\bigl(\tfrac{2x}{x+1}\bigr):
  1. \frac{d}{dx}\operatorname{sen}(x)=\cos(x).
  2. Para \frac{d}{dx}\!\bigl[-\cos(u)\bigr]=-(-\sin(u))\,u'=\sin(u)\,u', com u=\tfrac{2x}{x+1}.
    Calculamos u' pela regra do quociente:
u'=\frac{d}{dx}\frac{2x}{x+1} =\frac{2(x+1)-2x}{(x+1)^{2}} =\frac{2}{(x+1)^{2}}.
Logo
\frac{d}{dx}\Bigl[-\cos\bigl(\tfrac{2x}{x+1}\bigr)\Bigr] =\sin\!\Bigl(\tfrac{2x}{x+1}\Bigr)\,\frac{2}{(x+1)^{2}}.

Passo 3

Derivada de \tan(x^{2}) e de \operatorname{cotg}(x):
  1. Pela cadeia, \frac{d}{dx}\tan(x^{2})=\operatorname{sec}^{2}(x^{2})\cdot2x.
  2. \frac{d}{dx}\operatorname{cotg}(x)=-\csc^{2}(x).

Passo 4

Derivada de -\operatorname{sec}\bigl(\tfrac{x}{3}\bigr) e de \cossec(2\pi):
  1. Para v=\tfrac{x}{3}, \frac{d}{dx}\operatorname{sec}(v)=\operatorname{sec}(v)\tan(v)\,v'. Como v'=1/3,
\frac{d}{dx}\Bigl[-\operatorname{sec}\bigl(\tfrac{x}{3}\bigr)\Bigr] =-\operatorname{sec}\bigl(\tfrac{x}{3}\bigr)\tan\bigl(\tfrac{x}{3}\bigr)\,\frac{1}{3}.
  1. \cossec(2\pi) é constante (não depende de x), logo sua derivada é 0.

Passo 5

Derivada de -\operatorname{arctg}(2x): pela regra da cadeia, se w=2x então \frac{d}{dx}\operatorname{arctg}(w)=\frac{w'}{1+w^{2}}. Como w'=2,
\frac{d}{dx}\Bigl[-\operatorname{arctg}(2x)\Bigr] =-\frac{2}{1+(2x)^{2}} =-\frac{2}{1+4x^{2}}.
Agora somamos todas as parcelas de y'.

Resposta

y'=\cos(x) +\sin\!\Bigl(\tfrac{2x}{x+1}\Bigr)\,\frac{2}{(x+1)^{2}} +2x\,\operatorname{sec}^{2}(x^{2}) -\csc^{2}(x) -\frac{1}{3}\,\operatorname{sec}\!\Bigl(\tfrac{x}{3}\Bigr)\tan\!\Bigl(\tfrac{x}{3}\Bigr) -\frac{2}{1+4x^{2}}.

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