Determine as derivadas das funçocs.
g(i)=\left(\frac{1+\operatorname{sen} 3 i}{3-2 i}\right)^{-1}

Passo 1

O enunciado pede a derivada de
g(i)=\biggl(\frac{1+\operatorname{sen}(3i)}{3-2i}\biggr)^{-1}.
O conteúdo envolve derivada de função composta (regra da cadeia) e derivada de quociente.

Passo 2

Defina
h(i)=\frac{1+\operatorname{sen}(3i)}{3-2i},
de modo que g(i)=h(i)^{-1}. Pela regra da cadeia,
g'(i)=-\,h'(i)\,h(i)^{-2}.

Passo 3

Para h'(i) usamos a regra do quociente com \;u(i)=1+\operatorname{sen}(3i), \;v(i)=3-2i.
Então
u'(i)=3\cos(3i), \quad v'(i)=-2,
e
h'(i) =\frac{u'(i)\,v(i)-u(i)\,v'(i)}{[v(i)]^2} =\frac{3\cos(3i)\,(3-2i)-\bigl(1+\operatorname{sen}(3i)\bigr)(-2)}{(3-2i)^2} =\frac{3\cos(3i)\,(3-2i)+2\bigl(1+\operatorname{sen}(3i)\bigr)}{(3-2i)^2}.

Passo 4

Substituindo em
g'(i)=-\frac{h'(i)}{[h(i)]^2}
e lembrando que
[h(i)]^2 =\frac{(1+\operatorname{sen}(3i))^2}{(3-2i)^2},
temos
g'(i) =-\frac{3\cos(3i)\,(3-2i)+2\bigl(1+\operatorname{sen}(3i)\bigr)}{(3-2i)^2} \cdot\frac{(3-2i)^2}{\bigl(1+\operatorname{sen}(3i)\bigr)^2} =-\frac{3\cos(3i)\,(3-2i)+2\bigl(1+\operatorname{sen}(3i)\bigr)}{\bigl(1+\operatorname{sen}(3i)\bigr)^2}.

Resposta

g'(i)=-\frac{3\cos(3i)\,(3-2i)+2\bigl(1+\operatorname{sen}(3i)\bigr)}{\bigl(1+\operatorname{sen}(3i)\bigr)^2}.

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