# Calculating <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="normal">d </mrow>

Calculating $\frac{\mathrm{d}}{\mathrm{d}x}{\int }_{0}^{x}{e}^{\left(kt\right)}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$
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According to the fundamental theorem of calculus, if $g:\left[0,x\right]\to \mathbb{R}$ is continuous, then the function $F:\left[0,x\right]\to \mathbb{R}$ defined by
$F\left(x\right):={\int }_{0}^{x}g\left(t\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$
is differentiable on $\left(0,x\right),$ and ${F}^{\prime }=f$ In this case, you have $g\left(t\right)={e}^{kt}f\left(t\right),$$g\left(t\right)={e}^{kt}f\left(t\right),$ hence ${F}^{\prime }\left(x\right)={e}^{kx}f\left(x\right).$\