Question

Encontre a segunda derivada.

Answer 1

Resposta:

$\displaystyle\boxed{\dfrac{\textrm{d}^2}{\textrm{d}x^2}\int\limits_0^x \left(\int\limits_1^{\sin t} \sqrt{1+u^4}\textrm{ d}u\right)\textrm{ d}t = \cos x \sqrt{1+\sin^4 x}}.$

Explicação passo-a-passo:

Seja $\displaystyle f(\omega) = \int\limits_1^{\sin \omega} \sqrt{1+u^4}\textrm{ d}u.$

Então, tem-se:

$\displaystyle\int\limits_0^x \left(\int\limits_1^{\sin t} \sqrt{1+u^4}\textrm{ d}u\right)\textrm{ d}t = \int\limits_0^x f(t) \textrm{ d}t .$

Do teorema fundamental do cálculo, vem:

$\displaystyle\dfrac{\textrm{d}}{\textrm{d}x}\int\limits_0^x f(t) \textrm{ d}t = f(x),$

pelo que:

$\displaystyle\dfrac{\textrm{d}^2}{\textrm{d}x^2}\int\limits_0^x f(t) \textrm{ d}t = \dfrac{\textrm{d}f(x)}{\textrm{d}x}.$

Definindo agora:

$\displaystyle g(y) = \int\limits_1^{y} \sqrt{1+u^4} \textrm{ d}u,$

concluímos que $f(x) = g(\sin x)$, pelo que, da regra da cadeia:

$\dfrac{\textrm{d}f(x)}{\textrm{d}x} =\dfrac{\textrm{d}[g(\sin x)]}{\textrm{d}x} =\dfrac{\textrm{d}g(y)}{\textrm{d}y}(\sin x)\dfrac{\textrm{d}}{\textrm{d}x}(\sin x) = \cos x \dfrac{\textrm{d}g(y)}{\textrm{d}y}(\sin x).$

Aplicando de novo o teorema fundamental do cálculo, vem:

$\displaystyle \dfrac{\textrm{d}g(y)}{\textrm{d}y} = \displaystyle \dfrac{\textrm{d}}{\textrm{d}y}\int\limits_1^{y} \sqrt{1+u^4} \textrm{ d}u = \sqrt{1+y^4}.$

Portanto, tem-se por fim:

$\displaystyle\boxed{\dfrac{\textrm{d}^2}{\textrm{d}x^2}\int\limits_0^x \left(\int\limits_1^{\sin t} \sqrt{1+u^4}\textrm{ d}u\right)\textrm{ d}t = \cos x \sqrt{1+\sin^4 x}}.$