Determine a derivada da função dada:
f(x)=\arctan \left(\frac{5 \tan (x / 2)+4}{3}\right);

Passo 1

O enunciado pede a derivada de
f(x)=\arctan\Bigl(\frac{5\tan\bigl(\tfrac{x}{2}\bigr)+4}{3}\Bigr).
O conteúdo envolve regra da cadeia e as derivadas de \arctan(u) e de \tan(v).

Passo 2

Definimos
g(x)=\frac{5\tan\bigl(\frac{x}{2}\bigr)+4}{3},
de modo que f(x)=\arctan\bigl(g(x)\bigr). Pela regra da cadeia,
f'(x)=\frac{g'(x)}{1+\bigl(g(x)\bigr)^2}.

Passo 3

Para achar g'(x), escrevemos
g(x)=\frac{5}{3}\,\tan\Bigl(\frac{x}{2}\Bigr)+\frac{4}{3}.
Como
\frac{d}{dx}\tan\Bigl(\frac{x}{2}\Bigr) =\operatorname{sec}^2\Bigl(\frac{x}{2}\Bigr)\cdot\frac{1}{2},
vem
g'(x) =\frac{5}{3}\cdot\operatorname{sec}^2\Bigl(\frac{x}{2}\Bigr)\cdot\frac{1}{2} =\frac{5}{6}\,\operatorname{sec}^2\Bigl(\frac{x}{2}\Bigr).

Passo 4

Substituindo em f'(x), obtemos
f'(x) =\frac{\displaystyle\frac{5}{6}\,\operatorname{sec}^2\!\Bigl(\frac{x}{2}\Bigr)} {1+\displaystyle\Bigl(\frac{5\tan(\frac{x}{2})+4}{3}\Bigr)^2}.

Passo 5

Para simplificar o denominador, note que
1+\Bigl(\tfrac{5\tan(\frac{x}{2})+4}{3}\Bigr)^2 =\frac{9+(5\tan(\frac{x}{2})+4)^2}{9}.
Logo
f'(x) =\frac{5}{6}\operatorname{sec}^2\Bigl(\frac{x}{2}\Bigr)\cdot\frac{9}{9+(5\tan(\frac{x}{2})+4)^2} =\frac{\tfrac{15}{2}\,\operatorname{sec}^2\bigl(\tfrac{x}{2}\bigr)} {9+\bigl(5\tan(\tfrac{x}{2})+4\bigr)^2}.

Resposta

f'(x) =\frac{\frac{15}{2}\,\operatorname{sec}^2\!\bigl(\tfrac{x}{2}\bigr)} {9+\bigl(5\tan\bigl(\tfrac{x}{2}\bigr)+4\bigr)^2}.

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