Question

Find dy/dx and d^2*y/dx^2 if x=integral of (sin u)/u du from 1 to t and y=integral of e^u du from 2 to ln t.

Answer 1

\[x=\int\limits_{1}^{t}\frac{ \sin u }{ u }du\]

Answer 2

\[y=\int\limits_{2}^{\ln t}e^u du\]

Answer 3

@zepdrix

Answer 4

\[
\frac{dy}{dx} = \frac{dy}{dt} * \frac{dt}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}
\]

Answer 5

\[
y=\int\limits_{2}^{\ln t}e^u du \\
\frac{dy}{dt} = e^{\ln(t)} * \frac{d}{dt} \ln(t) = \frac{e^{\ln(t)} } { t } =
\frac{t} { t } = 1\\
\text{ } \\
x=\int\limits_{1}^{t}\frac{ \sin (u) }{ u }du \\
\frac{dx}{dt} = \frac{ \sin(t) }{ t } \\
\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} ~~~
\text{(provided } \frac{dx}{dt} \ne 0) \\
\frac{dy}{dx}=
1 \div \frac{\sin(t) }{t} \\
\frac{dy}{dx}= \frac{t }{\sin(t)} \\
\]

Answer 6

\[
\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) =
\frac{d}{dt}\left(\frac{dy}{dx}\right) * \frac{dt}{dx} =
\frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt} = ?
\]

Answer 7

Thanks a lot!