Use a regra da cadeia para encontrar \frac{d y}{d x} :
\left\{\begin{array}{l}x=\operatorname{arccos} \frac{1}{\sqrt{1-t^{2}}}} \\\ y=\operatorname{arcsen} \frac{1}{\sqrt{1+t^{2}}}\end{array}\right..

Passo 1

O enunciado define duas funções em parâmetro t:
x(t)=\operatorname{arccos}\bigl(u(t)\bigr) \quad\text{com}\quad u(t)=\frac{1}{\sqrt{1-t^{2}}},
y(t)=\operatorname{arcsen}\bigl(v(t)\bigr) \quad\text{com}\quad v(t)=\frac{1}{\sqrt{1+t^{2}}}.
Queremos \frac{dy}{dx}. Como y e x dependem de t, usamos
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.
Para isso vamos calcular \frac{dx}{dt} e \frac{dy}{dt} pela regra da cadeia, usando as derivadas
\frac{d}{du}\bigl(\operatorname{arccos} u\bigr)=-\frac{1}{\sqrt{1-u^{2}}} \quad\text{e}\quad \frac{d}{dv}\bigl(\operatorname{arcsen} v\bigr)=\frac{1}{\sqrt{1-v^{2}}}.

Passo 2

Calcular \frac{dx}{dt}. Temos
u(t)=(1-t^{2})^{-\tfrac12},
\frac{du}{dt} =-\tfrac12\,(1-t^{2})^{-\tfrac32}\cdot(-2t) =t\,(1-t^{2})^{-\tfrac32}.
Então
\frac{dx}{dt} =-\frac{1}{\sqrt{1-u^{2}}}\,\frac{du}{dt} =-\frac{1}{\sqrt{1-\frac{1}{1-t^{2}}}}\;t\,(1-t^{2})^{-\tfrac32}.
Note que
1-u^{2} =1-\frac{1}{1-t^{2}} =-\frac{t^{2}}{1-t^{2}}, \quad \sqrt{1-u^{2}}=\frac{t}{\sqrt{1-t^{2}}},
logo
\frac{dx}{dt} =-\frac{1}{\frac{t}{\sqrt{1-t^{2}}}}\;t\,(1-t^{2})^{-\tfrac32} =-\frac{\sqrt{1-t^{2}}}{t}\;t\,(1-t^{2})^{-\tfrac32} =-\frac{1}{1-t^{2}}.

Passo 3

Calcular \frac{dy}{dt}. Temos
v(t)=(1+t^{2})^{-\tfrac12},
\frac{dv}{dt} =-\tfrac12\,(1+t^{2})^{-\tfrac32}\cdot(2t) =-t\,(1+t^{2})^{-\tfrac32}.
Então
\frac{dy}{dt} =\frac{1}{\sqrt{1-v^{2}}}\,\frac{dv}{dt} =\frac{1}{\sqrt{1-\frac{1}{1+t^{2}}}}\;\bigl(-t\,(1+t^{2})^{-\tfrac32}\bigr).
Como
1-v^{2} =\frac{t^{2}}{1+t^{2}}, \quad \sqrt{1-v^{2}}=\frac{t}{\sqrt{1+t^{2}}},
segue
\frac{dy}{dt} =\frac{1}{\frac{t}{\sqrt{1+t^{2}}}}\;(-t\,(1+t^{2})^{-\tfrac32}) =-\frac{\sqrt{1+t^{2}}}{t}\;t\,(1+t^{2})^{-\tfrac32} =-\frac{1}{1+t^{2}}.

Passo 4

Agora
\frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{-\tfrac{1}{1+t^{2}}}{-\tfrac{1}{1-t^{2}}} =\frac{1-t^{2}}{1+t^{2}}.

Resposta

\frac{dy}{dx}=\frac{1-t^{2}}{1+t^{2}}.

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