Usando derivação implícita, calcule \mathrm{d} y / \mathrm{d} \alpha para a curva abaixo:
y-\operatorname{tg}^{2}(\operatorname{sen} 2 \alpha)+y(\cos 2 \alpha)-5=0

Passo 1

O enunciado dá a curva definida implicitamente por
y \;-\;\tan^{2}\bigl(\sin2\alpha\bigr)\;+\;y\cos2\alpha\;-\;5 \;=\;0
e pede \dfrac{\mathrm{d}y}{\mathrm{d}\alpha). O conteúdo é derivação implícita de uma equação que envolve y=y(\alpha) e funções de \alpha. Vamos diferenciar ambos os lados em relação a \alpha aplicando a regra da soma, produto e cadeia.

Passo 2

Diferenciando termo a termo:
  1. \dfrac{\mathrm{d}}{\mathrm{d}\alpha}[\,y\,]=\dfrac{\mathrm{d}y}{\mathrm{d}\alpha}.
  2. \dfrac{\mathrm{d}}{\mathrm{d}\alpha}\bigl[y\cos2\alpha\bigr] =\dfrac{\mathrm{d}y}{\mathrm{d}\alpha}\cos2\alpha \;+\;y\bigl(-\sin2\alpha\cdot2\bigr) =\cos2\alpha\,\dfrac{\mathrm{d}y}{\mathrm{d}\alpha}-2y\sin2\alpha.
  3. \dfrac{\mathrm{d}}{\mathrm{d}\alpha}\bigl[-\tan^{2}(\sin2\alpha)\bigr] =-2\,\tan\bigl(\sin2\alpha\bigr)\,\operatorname{sec}^{2}\bigl(\sin2\alpha\bigr)\cdot\dfrac{\mathrm{d}}{\mathrm{d}\alpha}[\sin2\alpha]
    e \dfrac{\mathrm{d}}{\mathrm{d}\alpha}[\sin2\alpha]=2\cos2\alpha.
    Logo
\dfrac{\mathrm{d}}{\mathrm{d}\alpha}\bigl[-\tan^{2}(\sin2\alpha)\bigr] =-2\,\tan(\sin2\alpha)\,\operatorname{sec}^{2}(\sin2\alpha)\,(2\cos2\alpha) =-4\cos2\alpha\;\tan(\sin2\alpha)\,\operatorname{sec}^{2}(\sin2\alpha).
  1. \dfrac{\mathrm{d}}{\mathrm{d}\alpha}[-5]=0. Somando tudo e igualando a zero, fica
\dfrac{\mathrm{d}y}{\mathrm{d}\alpha} +\cos2\alpha\,\dfrac{\mathrm{d}y}{\mathrm{d}\alpha} -2y\sin2\alpha -4\cos2\alpha\,\tan(\sin2\alpha)\,\operatorname{sec}^{2}(\sin2\alpha) =0.

Passo 3

Agrupando os termos que contêm \dfrac{\mathrm{d}y}{\mathrm{d}\alpha}:
\bigl(1+\cos2\alpha\bigr)\,\dfrac{\mathrm{d}y}{\mathrm{d}\alpha} -2y\sin2\alpha -4\cos2\alpha\,\tan(\sin2\alpha)\,\operatorname{sec}^{2}(\sin2\alpha) =0.
Isolando \dfrac{\mathrm{d}y}{\mathrm{d}\alpha}, obtemos
\bigl(1+\cos2\alpha\bigr)\,\dfrac{\mathrm{d}y}{\mathrm{d}\alpha} =2y\sin2\alpha +4\cos2\alpha\,\tan(\sin2\alpha)\,\operatorname{sec}^{2}(\sin2\alpha).

Passo 4

Dividindo por 1+\cos2\alpha, chegamos à expressão final:
\dfrac{\mathrm{d}y}{\mathrm{d}\alpha} =\frac{2y\sin2\alpha +4\cos2\alpha\,\tan(\sin2\alpha)\,\operatorname{sec}^{2}(\sin2\alpha)} {1+\cos2\alpha}.

Resposta

\dfrac{\mathrm{d}y}{\mathrm{d}\alpha} =\frac{2y\sin2\alpha +4\cos2\alpha\,\tan(\sin2\alpha)\,\operatorname{sec}^{2}(\sin2\alpha)} {1+\cos2\alpha}.

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