Às vezes a gente se vê perplexo ao descobrir que a expressão a ser derivada é complicada demais para se lidar diretamente.
Assim, a equação \[y = (x^2+a^2)^{\frac{3}{2}}\] é desajeitada para um iniciante.
Agora o artifício para contornar a dificuldade é este: Escreva algum símbolo, tal como \(u\), para a expressão \(x^2 + a^2\); então a equação fica \[y = u^{\frac{3}{2}},\] que você consegue lidar facilmente; pois \[\frac{dy}{du} = \frac{3}{2} u^{\frac{1}{2}}.\] Depois lide com a expressão \[u = x^2 + a^2,\] e derive-a com respeito a \(x\), \[\frac{du}{dx} = 2x.\] Então tudo que resta é simples;
pois \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx};\] ou seja, \[\begin{align} \frac{dy}{dx} &= \frac{3}{2} u^{\frac{1}{2}} \times 2x \\ &= \tfrac{3}{2} (x^2 + a^2)^{\frac{1}{2}} \times 2x \\ &= 3x(x^2 + a^2)^{\frac{1}{2}}; \end{align}\] e assim o truque está feito.
Com o tempo, quando você tiver aprendido como lidar com senos, cossenos e exponenciais, você verá a regra da cadeia de utilidade crescente.
Exemplos
Vamos praticar usando a Regra da Cadeia em alguns exemplos.
Exemplo 9.1. Derive \(y = \sqrt{a+x}\).
Solução. Seja \(a+x = u\). \[ \frac{du}{dx} = 1;\quad y=u^{\frac{1}{2}};\quad \frac{dy}{du} = \tfrac{1}{2} u^{-\frac{1}{2}} = \tfrac{1}{2} (a+x)^{-\frac{1}{2}}.\] \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{2\sqrt{a+x}}. \]
Exemplo 9.2. Derive \(y = \dfrac{1}{\sqrt{a+x^2}}\).
Solução. Seja \(a + x^2 = u\). \[ \frac{du}{dx} = 2x;\quad y=u^{-\frac{1}{2}};\quad \frac{dy}{du} = -\tfrac{1}{2}u^{-\frac{3}{2}}.\] \[\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx} = - \frac{x}{\sqrt{(a+x^2)^3}}. \]
Exemplo 9.3. Derive \(y = \left(m - nx^{\frac{2}{3}} + \dfrac{p}{x^{\frac{4}{3}}}\right)^a\).
Solução. Seja \(m - nx^{\frac{2}{3}} + px^{-\frac{4}{3}} = u\). \[ \frac{du}{dx} = -\tfrac{2}{3} nx^{-\frac{1}{3}} - \tfrac{4}{3} px^{-\frac{7}{3}};\] \[y = u^a;\quad \frac{dy}{du} = a u^{a-1}. \] \[\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx} = -a\left(m -nx^{\frac{2}{3}} + \frac{p}{x^{\frac{4}{3}}}\right)^{a-1} (\tfrac{2}{3} nx^{-\frac{1}{3}} + \tfrac{4}{3} px^{-\frac{7}{3}}). \]
Exemplo 9.4. Derive \(y=\dfrac{1}{\sqrt{x^3 - a^2}}\).
Solução. Seja \(u = x^3 - a^2\). \[\begin{align} \frac{du}{dx} &= 3x^2;\quad y = u^{-\frac{1}{2}};\quad \frac{dy}{du}=-\frac{1}{2}(x^3 - a^2)^{-\frac{3}{2}}. \\ \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} = -\frac{3x^2}{2\sqrt{(x^3 - a^2)^3}}. \end{align}\]
Exemplo 9.5. Derive \(y=\sqrt{\dfrac{1-x}{1+x}}\).
Solução. Escreva isto como \(y=\dfrac{(1-x)^{\frac{1}{2}}}{(1+x)^{\frac{1}{2}}}\). \[\frac{dy}{dx} = \frac{(1+x)^{\frac{1}{2}}\, \dfrac{d(1-x)^{\frac{1}{2}}}{dx} - (1-x)^{\frac{1}{2}}\, \dfrac{d(1+x)^{\frac{1}{2}}}{dx}}{1+x}.\]
(Também podemos escrever \(y = (1-x)^{\frac{1}{2}} (1+x)^{-\frac{1}{2}}\) e diferenciar como um produto.)
Procedendo como no exemplo 9.1 acima, obtemos \[\frac{d(1-x)^{\frac{1}{2}}}{dx} = -\frac{1}{2\sqrt{1-x}}; \quad\text{e}\quad \frac{d(1+x)^{\frac{1}{2}}}{dx} = \frac{1}{2\sqrt{1+x}}.\]
Portanto \[\begin{align} \frac{dy}{dx} &= - \frac{(1 + x)^{\frac{1}{2}}}{2(1 + x)\sqrt{1-x}} - \frac{(1 - x)^{\frac{1}{2}}}{2(1 + x)\sqrt{1+x}} \\ &= - \frac{1}{2\sqrt{1+x}\sqrt{1-x}} - \frac{\sqrt{1-x}}{2 \sqrt{(1+x)^3}};\\ \end{align}\] ou \[\frac{dy}{dx} = - \frac{1}{(1+x)\sqrt{1-x^2}}.\]
Exemplo 9.6. Derive \(y = \sqrt{\dfrac{x^3}{1+x^2}}\).
Solução. Podemos escrever isto \[\begin{gathered} y = x^{\frac{3}{2}}(1+x^2)^{-\frac{1}{2}}; \\ \frac{dy}{dx} = \tfrac{3}{2} x^{\frac{1}{2}}(1 + x^2)^{-\frac{1}{2}} + x^{\frac{3}{2}} \times \frac{d\bigl[(1+x^2)^{-\frac{1}{2}}\bigr]}{dx}. \end{gathered}\]
Diferenciando \((1+x^2)^{-\frac{1}{2}}\), como mostrado no exemplo (2) acima, obtemos \[\frac{d\bigl[(1+x^2)^{-\frac{1}{2}}\bigr]}{dx} = - \frac{x}{\sqrt{(1+x^2)^3}};\] de modo que \[\frac{dy}{dx} = \frac{3\sqrt{x}}{2\sqrt{1+x^2}} - \frac{\sqrt{x^5}}{\sqrt{(1+x^2)^3}} = \frac{\sqrt{x}(3+x^2)}{2\sqrt{(1+x^2)^3}}.\]
Exemplo 9.7. Derive \(y=\left(x+\sqrt{x^2+x+a}\right)^3\).
Solução. Seja \(x+\sqrt{x^2+x+a}=u\). \[\begin{gathered} \frac{du}{dx} = 1 + \frac{d\bigl[(x^2+x+a)^{\frac{1}{2}}\bigr]}{dx}. \\ y = u^3;\quad\text{e}\quad \frac{dy}{du} = 3u^2= 3\left(x+\sqrt{x^2+x+a}\right)^2. \end{gathered}\]
Agora seja \((x^2+x+a)^{\frac{1}{2}}=v\) e \((x^2+x+a) = w\). \[\begin{align} \frac{dw}{dx} &= 2x+1;\quad v = w^{\frac{1}{2}};\quad \frac{dv}{dw} = \tfrac{1}{2}w^{-\frac{1}{2}}. \\ \frac{dv}{dx} &= \frac{dv}{dw} \times \frac{dw}{dx} = \tfrac{1}{2}(x^2+x+a)^{-\frac{1}{2}}(2x+1). \end{align}\] Portanto \[\begin{align} \frac{du}{dx} &= 1 + \frac{2x+1}{2\sqrt{x^2+x+a}}, \\ \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx}\\ &= 3\left(x+\sqrt{x^2+x+a}\right)^2 \left(1 +\frac{2x+1}{2\sqrt{x^2+x+a}}\right). \end{align}\]
Exemplo 9.8. Derive \(y=\sqrt{\dfrac{a^2+x^2}{a^2-x^2}} \sqrt[3]{\dfrac{a^2-x^2}{a^2+x^2}}\).
Solução. Obtemos \[ y = \frac{(a^2+x^2)^{\frac{1}{2}} (a^2-x^2)^{\frac{1}{3}}} {(a^2-x^2)^{\frac{1}{2}} (a^2+x^2)^{\frac{1}{3}}} = (a^2+x^2)^{\frac{1}{6}} (a^2-x^2)^{-\frac{1}{6}}. \] \[\frac{dy}{dx} = (a^2+x^2)^{\frac{1}{6}} \frac{d\bigl[(a^2-x^2)^{-\frac{1}{6}}\bigr]}{dx} + \frac{d\bigl[(a^2+x^2)^{\frac{1}{6}}\bigr]}{(a^2-x^2)^{\frac{1}{6}}\, dx}. \]
Seja \(u = (a^2-x^2)^{-\frac{1}{6}}\) e \(v = (a^2 - x^2)\). \[ u = v^{-\frac{1}{6}};\quad \frac{du}{dv} = -\frac{1}{6}v^{-\frac{7}{6}};\quad \frac{dv}{dx} = -2x.\] \[\frac{du}{dx} = \frac{du}{dv} \times \frac{dv}{dx} = \frac{1}{3}x(a^2-x^2)^{-\frac{7}{6}}. \]
Seja \(w = (a^2 + x^2)^{\frac{1}{6}}\) e \(z = (a^2 + x^2)\). \[ w = z^{\frac{1}{6}};\quad \frac{dw}{dz} = \frac{1}{6}z^{-\frac{5}{6}};\quad \frac{dz}{dx} = 2x. \] \[\frac{dw}{dx} = \frac{dw}{dz} \times \frac{dz}{dx} = \frac{1}{3} x(a^2 + x^2)^{-\frac{5}{6}}. \]
Portanto \[ \frac{dy}{dx} = (a^2+x^2)^{\frac{1}{6}} \frac{x}{3(a^2-x^2)^{\frac{7}{6}}} + \frac{x}{3(a^2-x^2)^{\frac{1}{6}} (a^2+x^2)^{\frac{5}{6}}}; \] ou \[ \frac{dy}{dx} = \frac{x}{3} \left[\sqrt[6]{\frac{a^2+x^2}{(a^2-x^2)^7}} + \frac{1}{\sqrt[6]{(a^2-x^2)(a^2+x^2)^5]}} \right]. \]
Exemplo 9.9. Derive \(y^n\) com respeito a \(y^5\).
Solução.
\[\frac{d(y^n)}{d(y^5)} =\frac{\dfrac{d(y^n)}{dy}}{\dfrac{d(y^5)}{dy}}= \frac{ny^{n-1}}{5y^{5-1}} = \frac{n}{5} y^{n-5}.\]
Exemplo 9.10. Encontre a primeira e segunda derivadas de \(y = \dfrac{x}{b} \sqrt{(a-x)x}\).
Solução. \[\frac{dy}{dx} = \frac{x}{b}\, \frac{d\bigl\{\bigl[(a-x)x\bigr]^{\frac{1}{2}}\bigr\}}{dx} + \frac{\sqrt{(a-x)x}}{b}.\tag{Regra do Produto}\]
Seja \(\bigl[(a-x)x\bigr]^{\frac{1}{2}} = u\) e seja \((a-x)x = w\); então \(u = w^{\frac{1}{2}}\). \[\frac{du}{dw} = \frac{1}{2} w^{-\frac{1}{2}} = \frac{1}{2w^{\frac{1}{2}}} = \frac{1}{2\sqrt{(a-x)x}}.\] \[\begin{align} &\frac{dw}{dx} = a-2x.\\ &\frac{du}{dw} \times \frac{dw}{dx} = \frac{du}{dx} = \frac{a-2x}{2\sqrt{(a-x)x}}. \end{align}\]
Portanto \[\frac{dy}{dx} = \frac{x(a-2x)}{2b\sqrt{(a-x)x}} + \frac{\sqrt{(a-x)x}}{b} = \frac{x(3a-4x)}{2b\sqrt{(a-x)x}}.\]
Agora \[\begin{align} \frac{d^2y}{dx^2} &= \frac{2b \sqrt{(a-x)x}\, (3a-8x) - \dfrac{(3ax-4x^2)b(a-2x)}{\sqrt{(a-x)x}}} {4b^2(a-x)x} \\ &= \frac{3a^2-12ax+8x^2}{4b(a-x)\sqrt{(a-x)x}}. \end{align}\]
(Precisaremos dessas duas últimas derivadas mais adiante. Veja o exercício 11 do Capítulo 12.)
Exemplo 9.11. Um cilindro cuja altura é duas vezes o raio da base está aumentando em volume, de modo que todas as suas partes mantêm sempre a mesma proporção umas com as outras; ou seja, em qualquer instante, o cilindro é similar ao cilindro original. Quando o raio da base é \(r\) feet, a área da superfície está aumentando à taxa de \(20\) square inches por segundo; a qual taxa seu volume está aumentando então?1
Solução. \[\text{Área}= S = 2(\pi r^2)+ 2 \pi r \times 2r = 6 \pi r^2.\] \[\text{Volume} = V = \pi r^2 \times 2r=2 \pi r^3.\] \[\frac{dS}{dt} = 12\pi r\dfrac{dr}{dt}=20,\quad\Rightarrow\quad \frac{dr}{dt}=\frac{20}{12 \pi r},\] \[\frac{dV}{dt} = 6\pi r^2\, \frac{dr}{dt} = 6 \pi r^2 \times \frac{20}{12 \pi r} = 10r. \]
O volume muda à taxa de \(10r\) cubic inches.
Exercícios I
Derive o seguinte:
Exercício 9.1. \(y = \sqrt{x^2 + 1}\).
Resposta
\(\dfrac{x}{\sqrt{ x^2 + 1}}\).
Solução
Seja \(u=x^2+1\). Então \(y=u^{\frac{1}{2}}\) e \[\begin{align} \frac{d y}{d x}&=\frac{dy}{du}\cdot\frac{du}{dx}\\ &=\frac{1}{2}u^{\frac{1}{2}-1}\cdot (2 x)\\ &=x\left(x^{2}+1\right)^{-1 / 2}\\ &=\frac{x}{\sqrt{x^{2}+1}} \end{align}\]
Exercício 9.2. \(y = \sqrt{x^2+a^2}\).
Resposta
\(\dfrac{x}{\sqrt{ x^2 + a^2}}\).
Solução
\[y=\sqrt{x^{2}+a^{2}}\]
Seja \(u=x^2+a^2\). Então \(y=u^{\frac{1}{2}}\), e
\[\begin{align} \frac{d y}{d x}&=\frac{dy}{du}\cdot\frac{du}{dx}\\ &=\left(\frac{1}{2}u^{-\frac{1}{2}}\right)\cdot (2x)\\ &=\frac{1}{2}\left(x^{2}+a^{2}\right)^{\frac{1}{2}-1}(2 x)\\ &=\frac{x}{\sqrt{x^{2}+a^{2}}} \end{align}\]
Exercício 9.3. \(y = \dfrac{1}{\sqrt{a+x}}\).
Resposta
\(- \dfrac{1}{2 \sqrt{(a + x)^3}}\).
Solução
\[\begin{align} & y=\frac{1}{\sqrt{a+x}}=(a+x)^{-1 / 2} \\ \Rightarrow &\frac{d y}{d x}=\left(-\frac{1}{ 2}\right)(a+x)^{-\frac{1}{2}-1}=\frac{-1}{2 \sqrt[2]{(a+x)^{3}}} \end{align}\]
Exercício 9.4. \(y = \dfrac{a}{\sqrt{a-x^2}}\).
Resposta
\(\dfrac{ax}{\sqrt{(a - x^2)^3}}\).
Solução
\[y=\frac{a}{\sqrt{a-x^{2}}}=a\left(a-x^{2}\right)^{-\frac{1}{2}}\]
Seja \(u=a-x^{2}\). Então
\[y=a u^{-\frac{1}{2}}\]
e
\[\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\left(-\frac{a}{2} u^{-\frac{3}{2}}\right)(-2 x) \\ & =\frac{a x}{u^{\frac{3}{2}}} \\ & =\frac{a x}{\sqrt{\left(a-x^{2}\right)^{3}}} \end{align}\]
Exercício 9.5. \(y = \dfrac{\sqrt{x^2-a^2}}{x^2}\).
Resposta
\(\dfrac{2a^2 - x^2}{x^3 \sqrt{ x^2 - a^2}}\).
Solução
\[y=\frac{\sqrt{x^{2}-a^{2}}}{x^{2}}\] Usando a Regra do Quociente
\[\frac{d y}{d x}=\frac{\frac{d\left(\sqrt{x^{2}-a^{2}}\right)}{d x} \cdot x^{2}-2 x \sqrt{x^{2}-a^{2}}}{x^{4}}\]
Para encontrar \(\frac{d u}{d x}\) onde \(u=\sqrt{x^2-a^2}\), seja \(v=x^{2}-a^{2}\), então \(u=\sqrt{v}\) e
\[\begin{align} \frac{d u}{d x} & =\frac{d u}{d v} \cdot \frac{d v}{d x} \\ & =\frac{1}{2 \sqrt{v}} \cdot(2 x) \\ & =\frac{x}{\sqrt{x^{2}-a^{2}}} . \end{align}\] Portanto \[\frac{d\left(\sqrt{x^{2}-a^{2}}\right)}{d x}=\frac{x}{\sqrt{x^{2}-a^{2}}}\] e \[\begin{align} \frac{d y}{d x} & =\frac{\dfrac{x}{\sqrt{x^{2}-a^{2}}} \cdot x^{2}-2 x \sqrt{x^{2}-a^{2}}}{x^{4}} \\ & =\frac{x^{3}-2 x\left(x^{2}-a^{2}\right)}{x^{4} \sqrt{x^{2}-a^{2}}} \\ & =\frac{2 a x-x^{3}}{x^{4} \sqrt{x^{2}-a^{2}}} \\ & =\frac{2 a-x^{2}}{x^{3} \sqrt{x^{2}-a^{2}}} \end{align}\]
Exercício 9.6. \(y = \dfrac{\sqrt[3]{x^4+a}}{\sqrt{x^3+a}}\).
Resposta
\(\dfrac{\frac{3}{2} x^2 \left[ \frac{8}{9} x \left( x^3 + a \right) - \left( x^4 + a \right) \right]}{(x^4 + a)^{\frac{2}{3}} (x^3 + a)^{\frac{3}{2}}}\)
Solução
Para encontrar \(\frac{d y}{d x}\), precisamos encontrar \(\frac{d\left(\sqrt[3]{x^{4}+a}\right)}{d x}\) e \(\frac{d\left(\sqrt{x^{3}+a}\right)}{d x}\).
\(u=\sqrt[3]{x^{4}+a}=\left(x^{4}+a\right)^{\frac{1}{3}}=v^{\frac{1}{3}}\) onde \(v=x^{4}+a\). Então
\[\begin{align} \frac{d u}{d x} & =\frac{d u}{d u} \cdot \frac{d u}{d x} \\ & =\left(\frac{1}{3} v^{-\frac{2}{3}}\right)\left(4 x^{3}\right) \\ & =\frac{4 x^{3}}{3 \sqrt{\left(x^{2}+a\right)^{3}}} \end{align}\]
\(w=\sqrt{x^{3}+a}=\left(x^{3}+a\right)^{\frac{1}{2}}=z^{\frac{1}{2}}\), onde \(z=x^{3}+a\). Então
\[\begin{align} \frac{d w}{d x} & =\frac{d w}{d z} \cdot \frac{d z}{d x} \\ & =\frac{1}{2} z^{-\frac{1}{2}} \cdot\left(3 x^{2}\right) \\ & =\frac{3 x^{2}}{2 z^{\frac{1}{2}}} \\ & =\frac{3 x^{2}}{2 \sqrt{x^{3}+a}} \end{align}\]
Agora usando a Regra do Quociente:
\[\begin{align} \frac{d y}{d x} & =\frac{\frac{d\left(\sqrt[3]{x^{4}+a}\right)}{d x} \cdot \sqrt{x^{3}+a}-\frac{d\left(\sqrt{x^{3}+a}\right)}{d x} \sqrt[3]{x^{4}+a}}{x^{3}+a} \\ & =\frac{\frac{4 x^{3}}{3 \sqrt{\left(x^{4}+a\right)^{3}}} \sqrt{x^{3}+a}-\frac{3 x^{2}}{2 \sqrt{x^{3}+a}} \cdot \sqrt[3]{x^{4}+a}}{x^{3}+a} \\ & =\frac{\frac{4}{3} x^{3}\left(x^{3}+a\right)-\frac{3}{2} x^{2}\left(x^{4}+a\right)}{\sqrt{\left(x^{4}+a\right)^{3}}\left(x^{3}+a\right)^{\frac{3}{2}}} \\ & =\frac{\frac{3}{2} x^{2}\left[\frac{8}{9} x\left(x^{3}+a\right)-\left(x^{4}+a\right)\right]}{\left(x^{4}+a\right)^{\frac{3}{2}}\left(x^{3}+a\right)^{\frac{3}{2}}}. \end{align}\]
Exercício 9.7. \(y = \dfrac{a^2+x^2}{(a+x)^2}\).
Resposta
\(\dfrac{2a \left(x - a \right)}{(x + a)^3}\).
Solução
Usando a Regra do Quociente: \[\begin{align} & \frac{d y}{d x}=\frac{2 x(a+x)^{2}-2(a+x)\left(a^{2}+x^{2}\right)}{(a+x)^{4}} \\ & \frac{d y}{d x}=\frac{2(a+x)\left[x(a+x)-\left(a^{2}+x^{2}\right)\right]}{(a+x)^{4}} \\ & \frac{d y}{d x}=\frac{2\left[a x+x^{2}-a^{2}-x^{2}\right]}{(a+x)^{3}}=\frac{2 a(x-a)}{(a+x)^{3}} \end{align}\]
Observe que para encontrar \(\dfrac{d\left((a+x)^2\right)}{dx}\), seja \(u=a+x\). Então \[\frac{d\left((a+x)^2\right)}{dx}=\frac{d (u^2)}{du}\frac{du}{dx}=2u=2(a+x)\]
Exercício 9.8. Derive \(y^5\) com respeito a \(y^2\).
Resposta
\(\frac{5}{2} y^3\).
Solução
\[\frac{d \left(y^{5}\right)}{d \left(y^{2}\right)}=\frac{\dfrac{d\left(y^{5}\right)}{d y}}{\dfrac{d\left(y^{2}\right)}{d y}}=\frac{5 y^{4}}{2 y}=\frac{5}{2} y^{3}\]
Exercício 9.9. Derive \(y = \dfrac{\sqrt{1 - \theta^2}}{1 - \theta}\).
Resposta
\(\dfrac{1}{(1 - \theta) \sqrt{1 - \theta^2}}\).
Solução
\[y=\frac{\sqrt{1-\theta^{2}}}{1-\theta}\]
Para encontrar \(\dfrac{dy}{d\theta}\), primeiro precisamos encontrar a derivada do numerador. Para diferenciar \(\sqrt{1-\theta^2}\) com respeito a \(\theta\), podemos reescrevê-la como \(\left(1 - \theta^2\right)^{\frac{1}{2}}\) e aplicar a Regra da Cadeia. Seja \(u = 1-\theta^2\). Então: \[\begin{align} \frac{d\left( \left(1 - \theta^2\right)^{\frac{1}{2}}\right)}{dx}&=\frac{d \left(u^\frac{1}{2}\right)}{d\theta}\\ &=\frac{d \left(u^\frac{1}{2}\right)}{du}\frac{u}{d\theta}\\ &=\frac{1}{2} u^{-\frac{1}{2}} (-2\theta)\\ &=\frac{1}{2}(1-\theta^2)^{-\frac{1}{2}}(-2\theta)\\ &=-\frac{\theta}{\sqrt{1-\theta^2}} \end{align}\] Usando a Regra do Quociente
\[\begin{align} \frac{d y}{d \theta}&=\frac{-\dfrac{\theta}{\sqrt{1-\theta^2}(1-\theta)}-(-1) \sqrt{1-\theta^{2}}}{(1-\theta)^{2}}\\ &=\frac{-\theta(1-\theta)+(1-\theta^2)}{\sqrt{1-\theta^2}\ \ (1-\theta^2)}\\ &=\frac{1-\theta}{\sqrt{1-\theta^2}\ \ (1-\theta)^2}\\ &=\frac{1}{\sqrt{1-\theta^2}\ \ (1-\theta)}. \end{align}\]
Exercício 9.10. Um balão esférico está aumentando em volume. Se, quando seu raio é \(r\) feet, seu volume está aumentando à taxa de \(4\) cubic feet por segundo, a qual taxa sua área de superfície está aumentando então?2
Resposta
À taxa de \(\dfrac{8}{r}\) square feet por segundo.
Solução
O volume do balão é
\[V=\frac{4}{3} \pi r^{3}\]
e a área da superfície do balão é
\[S=4 \pi r^{2}\]
Sabemos
\[\frac{d V}{d t}=4~\frac{\mathrm{ft}^{3}}{\mathrm{~s}}\]
Queremos encontrar \(\dfrac{d S}{d t}\).
Diferencie ambos os lados da seguinte equação com respeito ao tempo \(t\)
\[\frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}\]
Como \(\dfrac{d V}{d t}=4\), temos
\[\frac{d r}{d t}=\frac{1}{\pi r^{2}}\]
Agora diferencie ambos os lados de \(S=4 \pi r^{2}\), com respeito ao tempo \(t:\)
\[\frac{d S}{d t}=8 \pi r \frac{d r}{d t}\]
Substitua \(\frac{d r}{d t}=\frac{1}{\pi r^{2}}\) na equação acima obtém
\[\frac{d S}{d t}=8 \pi r \frac{1}{\pi r^{2}}=\frac{8}{r} .\]
O processo pode ser estendido a três ou mais derivadas, de modo que \(\dfrac{dy}{dx} = \dfrac{dy}{dz} \times \dfrac{dz}{dv} \times \dfrac{dv}{dx}\).
Exemplos
Exemplo 9.12. Se \(z = 3x^4\);\(v = \dfrac{7}{z^2}\);\(y =\sqrt{1+v}\), encontre \(\dfrac{dv}{dx}\).
Solução. Temos \[ \frac{dy}{dv} = \frac{1}{2\sqrt{1+v}};\quad \frac{dv}{dz} = -\frac{14}{z^3};\quad \frac{dz}{dx} = 12x^3. \] \[\frac{dy}{dx} = -\frac{168x^3}{(2\sqrt{1+v})z^3}= -\frac{28}{3x^5\sqrt{9x^8+7}}. \]
Exemplo 9.13. Se \(t = \dfrac{1}{5\sqrt{\theta}}\);\(x = t^3 + \dfrac{t}{2}\);\(v = \dfrac{7x^2}{\sqrt[3]{x-1}}\), encontre \(\dfrac{dv}{d\theta}\).
Solução. Como \[\frac{dv}{d\theta}=\frac{dv}{dx}\cdot\frac{dx}{dt}\cdot\frac{dt}{d\theta}\] para calcular \(\dfrac{dv}{d\theta}\), primeiro precisamos encontrar \(\dfrac{dv}{dx}\), \(\dfrac{dx}{dt}\), e \(\dfrac{dt}{d\theta}\).
Diferenciando \(v\) com respeito a \(x\) obtém \[\frac{dv}{dx}=\frac{\sqrt[3]{x-1}\dfrac{d(7x^{2})}{dx}-7x^{2}\dfrac{d(\sqrt[3]{x-1})}{dx}}{\left(\sqrt[3]{x-1}\right)^{2}} \tag{Regra do Quociente}\] Colocando \(\dfrac{d(7x^2)}{dx}=14x\) e \(\dfrac{d\left(\sqrt[3]{x-1})\right)}{dx}=\dfrac{d\left((x-1)^{\frac{1}{3}}\right)}{dx}=\frac{1}{3}(x-1)^{-\frac{2}{3}}\) na expressão acima, obtemos \[\frac{dv}{dx}=\frac{\sqrt[3]{x-1}(14x)-\frac{7}{3}x^{2}(x-1)^{-\frac{2}{3}}}{\left(\sqrt[3]{x-1}\right)^{2}}\] Para simplificar, multiplique tanto o numerador quanto o denominador por \(3(x-1)^{\frac{2}{3}}\): \[\begin{align} \frac{dv}{dx} &=\frac{7x\left[2(x-1)^{\frac{1}{3}}-\frac{1}{3}x(x-1)^{-\frac{2}{3}}\right]}{(x-1)^{\frac{2}{3}}}\times\frac{3(x-1)^{\frac{2}{3}}}{3(x-1)^{\frac{2}{3}}}\\[9pt] &=\frac{7x\left[6(x-1)-x\right]}{3(x-1)^{\frac{4}{3}}}\\ &=\frac{7x(5x-6)}{3\sqrt[3]{(x-1)^4}}. \end{align}\]
\[x = t^3 + \dfrac{t}{2}\quad\Rightarrow \quad \frac{dx}{dt}=3t^2+\frac{1}{2}\]
\[t = \dfrac{1}{5\sqrt{\theta}}=\frac{1}{5}\theta^{-\frac{1}{2}}\quad\Rightarrow\quad\dfrac{dt}{d\theta}=\frac{1}{5}\times\left(-\frac{1}{2}\right)\theta^{-\frac{3}{2}}=-\frac{1}{10\sqrt{\theta^3}}\]
Então \[\begin{gathered} \frac{dv}{dx} = \frac{7x(5x-6)}{3\sqrt[3]{(x-1)^4}};\quad \frac{dx}{dt} = 3t^2 + \tfrac{1}{2};\quad \frac{dt}{d\theta} = -\frac{1}{10\sqrt{\theta^3}}. \end{gathered}\] Portanto \[\begin{gathered} \frac{dv}{d\theta} = -\frac{7x(5x-6)(3t^2+\frac{1}{2})} {30\sqrt[3]{(x-1)^4} \sqrt{\theta^3}}, \end{gathered}\] uma expressão na qual \(x\) deve ser substituído por seu valor, e \(t\) por seu valor em termos de \(\theta\).
Exemplo 9.14. Se \(\theta = \dfrac{3a^2x}{\sqrt{x^3}}\);\(\omega = \dfrac{\sqrt{1-\theta^2}}{1+\theta}\);e \(\phi = \sqrt{3} - \dfrac{1}{\omega\sqrt{2}}\), encontre \(\dfrac{d\phi}{dx}\).
Solução. Obtemos \[\theta = 3a^2x^{-\frac{1}{2}};\quad \omega = \frac{\sqrt{(1-\theta)(1+\theta)}}{1+\theta}=\sqrt{\frac{1-\theta}{1+\theta}};\quad \text{e}\quad \phi = \sqrt{3} {-} \frac{1}{\sqrt{2}} \omega^{-1}.\] \[\frac{d\theta}{dx} = -\frac{3a^2}{2\sqrt{x^3}};\quad \frac{d\omega}{d\theta} = -\frac{1}{(1+\theta)\sqrt{1-\theta^2}}\] (veja exemplo 9.5); e \[\frac{d\phi}{d\omega} = \frac{1}{\sqrt{2}\omega^2}.\]
Portanto \(\dfrac{d\theta}{dx} = \dfrac{1}{\sqrt{2} \times \omega^2} \times \dfrac{1}{(1+\theta) \sqrt{1-\theta^2}} \times \dfrac{3a^2}{2\sqrt{x^3}}\).
Substitua agora primeiro \(\omega\), depois \(\theta\) por seu valor.
Exercícios II
Exercício 9.11. Se \(u = \frac{1}{2}x^3\);\(v = 3(u+u^2)\); e \(w = \dfrac{1}{v^2}\), encontre \(\dfrac{dw}{dx}\).
Resposta
\(\dfrac{dw}{dx} = -\dfrac{x^2 \left( 1 + x^3 \right)} {3 \left(\frac{1}{2} x^3 + \frac{1}{4} x^6 \right)^3}\).
Solução
\[\begin{align} & \frac{d w}{d x}=\frac{d w}{d v} \cdot \frac{d v}{d u} \cdot \frac{d u}{d x} \\ & =\frac{d\left(v^{-2}\right)}{d v} \cdot \frac{d\left(3 u+3 u^{2}\right)}{d u} \cdot \frac{d\left(\frac{1}{2} x^{3}\right)}{d x} \\ & =\left(-2 v^{-3}\right)(3+6 u)\left(\frac{3}{2} x^{2}\right) \\ & =\left(-\frac{2}{v^3}\right)(3+6 u)\left(\frac{3}{2} x^{2}\right) \\ & =\frac{-2}{\left(3 u+3 u^{2}\right)^{3}}(3+6 u)\left(\frac{3}{2} x^{2}\right) \\ & =\frac{-6(1+2 u)}{27\left(u+u^{2}\right)^{3}} \times \frac{3}{2} x^{2} \\ & =-\frac{1}{3} \frac{\left(1+2 \times \frac{1}{2} x^{3}\right)}{\left(\frac{1}{2} x^{3}+\frac{1}{4} x^{6}\right)^{3}} x^{2} \\ & =-\frac{x^{2}}{3} \frac{\left(1+x^{3}\right)}{\left(\frac{1}{2} x^{3}+\frac{1}{4} x^{6}\right)^{3}} \end{align}\]
Exercício 9.12. Se \(y = 3x^2 + \sqrt{2}\);\(z = \sqrt{1+y}\); e \(v = \dfrac{1}{\sqrt{3}+4z}\), encontre \(\dfrac{dv}{dx}\).
Resposta
\(\dfrac{dv}{dx} = - \dfrac{12x}{\sqrt{1 + \sqrt{2} + 3x^2} \left(\sqrt{3} + 4 \sqrt{1 + \sqrt{2} + 3x^2}\right)^2}\).
Solução
\[y=3 x^{2}+\sqrt{2}, \quad z=\sqrt{1+y},\quad v=\frac{1}{\sqrt{3}+4 z}.\]
\[\frac{d y}{d x}=6 x,\] \[\begin{align} \frac{d z}{d y}&=\frac{1}{2}(1+y)^{\frac{1}{2}-1}\\ &=\frac{1}{2 \sqrt{1+y}}\\ &=\frac{1}{2 \sqrt{1+3 x^{2}+\sqrt{2}}}\\ &=\frac{1}{2 \sqrt{1+\sqrt{2}+3 x^{2}}} \end{align}\] \[\begin{align} \frac{d v}{d z}&=\frac{-4}{(\sqrt{3}+4 z)^{2}}\\ &=\frac{-4}{(\sqrt{3}+4 \sqrt{1+y})^{2}}\\ &=\frac{-4}{\left(\sqrt{3}+4 \sqrt{1+\sqrt{2}+3 x^{2}}\right)^{2}} \end{align}\]
\[\begin{align} \frac{d v}{d x}&=\frac{d v}{d z} \cdot \frac{d z}{d y} \cdot \frac{d y}{d x}= \\ & =\frac{-4}{\left(\sqrt{3}+4 \sqrt{1+\sqrt{2}+3 x^{2}}\right)^{2}} \cdot \frac{1}{2 \sqrt{1+\sqrt{2}+3 x^{2}}} \cdot 6 x \\ & =\frac{-12 x}{\sqrt{1+\sqrt{2}+3 x^{2}}\left(\sqrt{3}+4 \sqrt{1+\sqrt{2}+3 x^{2}}\right)^{2}} \\ \end{align}\]
Exercício 9.13. Se \(y = \dfrac{x^3}{\sqrt{3}}\);\(z = (1+y)^2\); e \(u = \dfrac{1}{\sqrt{1+z}}\), encontre \(\dfrac{du}{dx}\).
Resposta
\(\dfrac{du}{dx} = - \dfrac{x^2 \left(\sqrt{3} + x^3 \right)} {\sqrt{ \left[ 1 + \left( 1 + \dfrac{x^3}{\sqrt{3}} \right) ^2 \right]^3}}\).
Solução
\[y =\frac{x^{3}}{\sqrt{3}} \Rightarrow \frac{d y}{d x}=\frac{3}{\sqrt{3}} x^{2}=\sqrt{3} x^{2}\]
\[z =(1+y)^{2} \Rightarrow \frac{d z}{d y}=2(1+y)\] \[u =\frac{1}{\sqrt{1+z}}=(1+z)^{-\frac{1}{2}} \Rightarrow \frac{d u}{d z}=-\frac{1}{2}(1+z)^{-\frac{3}{2}}\]
\[\begin{align} \frac{d u}{d x} & =\frac{d u}{d z} \cdot \frac{d z}{d y} \cdot \frac{d y}{d x} \\ & =-\frac{1}{2(1+z)^{\frac{3}{2}}} \cdot 2(1+y) \times \sqrt{3} x^{2} \\ & =-\frac{\sqrt{3} x^{2}}{\left(1+(1+y)^{2}\right)^{\frac{3}{2}}}(1+y) \\ & =-\frac{\sqrt{3} x^{2}}{\left(1+\left(1+\dfrac{x^{3}}{\sqrt{3}}\right)^{2}\right)^{\frac{3}{2}}}\left(\sqrt{3}+x^{3}\right) \\ & =-\frac{\sqrt{3} x^2}{\left(1+\left(1+\dfrac{x^3}{\sqrt{3}}\right)^2\right)^{\frac{3}{2}}}\left(1+\frac{x^3}{\sqrt{3}}\right) \\ & =-\frac{x^2\left(\sqrt{3}+x^3\right)}{\sqrt{\left[1+\left(1+\dfrac{x^3}{\sqrt{3}}\right)^2\right]^3}} \end{align}\]
No livro original, este era um exemplo no Capítulo 11.↩︎
No livro original, este era um exercício no Capítulo 11.↩︎
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