For derivatives of the matrix exponential there is a useful formula: $$ \frac{\mathrm{d}}{\mathrm{d}t} e^{M(t)} = \int^1_0 e^{\alpha M(t)} \frac{\mathrm{d}M(t)}{\mathrm{d}t} e^{(1-\alpha)M(t)}$$ Is there an extension to functional derivatives? $$ \frac{\delta}{\delta f(u)} e^{M[t, f]} = \int^1_0 e^{\alpha M[t,f]} \frac{\delta M[t,f]}{\delta f(u)} e^{(1-\alpha)M[t,f]}$$ If not, what about a simpler scenario where $M$ is simply a function of $f(t)$? $$ \frac{\delta}{\delta f(u)} e^{M(t,f(t))} = \delta(t-u)\int^1_0 e^{\alpha M(t,f(t))} \frac{\partial M(t,f(t))}{\partial f(t)} e^{(1-\alpha)M(t,f(t))}$$
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