# I have the following exepression in my book: d x </mrow>

I have the following exepression in my book:

Then it says, multiply both sides of the differential equation by the integrating factor $I\left(t\right)$.
$I\left(t\right)\frac{dx\left(t\right)}{dt}+{a}_{1}\left(t\right)I\left(t\right)x\left(t\right)=I\left(t\right)g\left(t\right)$
So far so good. Hereafter it says, the left-hand side is an exact derivative.
$\frac{d\left[x\left(t\right)I\left(t\right)\right]}{dt}=I\left(t\right)g\left(t\right)$
And my question is, how does the book come to the last? Can anyone give a HINT.
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lywiau63
I suppose that the integrating factor $I$ is defined by
$I\left(t\right)=\mathrm{exp}\left({\int }_{{t}_{0}}^{t}{a}_{1}\left(s\right)\phantom{\rule{thinmathspace}{0ex}}ds\right)$
and hence has the property
${I}^{\prime }\left(t\right)=\mathrm{exp}\left({\int }_{{t}_{0}}^{t}{a}_{1}\left(s\right)\phantom{\rule{thinmathspace}{0ex}}ds\right){a}_{1}\left(t\right)=I\left(t\right){a}_{1}\left(t\right)$
so
$\frac{d}{dt}\left(x\left(t\right)I\left(t\right)\right)=\frac{d}{dt}x\left(t\right)\cdot I\left(t\right)+x\left(t\right)\cdot \frac{d}{dt}I\left(t\right)=\frac{d}{dt}x\left(t\right)\cdot I\left(t\right)+x\left(t\right){a}_{1}\left(t\right)I\left(t\right)$
which is the left hand side.