Question

Se y=∫3xln(t)dt então dydx=3ln(x). QUESTÃO DE VERDADEIRO OU FALSO;

Answer 1

$\boxed{\int\limits^b_a{f(x)\mathrm{d}x}=G(b)-G(a)}$

$\boxed{\int{uv'}=uv-\int{u'v}}$

$y=\int\limits^3_x{(\ln{t})\mathrm{d}t}$

Podemos integrar por partes:

$u=\ln{t}\ \to\ u'=\dfrac{1}{t}$

$v'=1\ \to\ v=t$

$y=\bigg(t\ln{t}-\int{\mathrm{d}t}\bigg)\bigg|^3_x=\bigg(t\ln{t}-t\bigg)\bigg|^3_x$

Assim, avaliando os limites de integração:

$y=3\ln{3}-3-(x\ln{x}-x)$

$\boxed{y=x(1-\ln{x})+3(\ln{3}-1)}$

Dessa forma, a derivada de y em relação a x será:

$\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}}{\mathrm{d}x}\bigg(x(1-\ln{x})\bigg)+\dfrac{\mathrm{d}}{\mathrm{d}x}\bigg(3(\ln{3}-1)\bigg)$

$\dfrac{\mathrm{d}y}{\mathrm{d}x}=(1-\ln{x})\dfrac{\mathrm{d}}{\mathrm{d}x}(x)+x\dfrac{\mathrm{d}}{\mathrm{d}x}(1-\ln{x})+0$

$\dfrac{\mathrm{d}y}{\mathrm{d}x}=(1-\ln{x})(1)+x\bigg(0-\dfrac{1}{x}\bigg)$

$\dfrac{\mathrm{d}y}{\mathrm{d}x}=1-\ln{x}-1$

$\boxed{\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\ln{x}}$

Falso.