If y=e^x, dy/dx=e^x

\underset { \delta x\rightarrow 0 }{ lim } \frac { f\left( x+\delta x \right) -f\left( x \right) }{ \delta x } \\ \\ =\underset { \delta x\rightarrow 0 }{ lim } \frac { { e }^{ x+\delta x }-{ e }^{ x } }{ \delta x } \\ \\ =\underset { \delta x\rightarrow 0 }{ lim } \frac { { e }^{ x }{ e }^{ \delta x }-{ e }^{ x } }{ \delta x } \\ \\ =\underset { \delta x\rightarrow 0 }{ lim } \frac { { e }^{ x }\left( { e }^{ \delta x }-1 \right) }{ \delta x } \\ \\ ={ e }^{ x }\underset { \delta x\rightarrow 0 }{ lim } \frac { \left( { e }^{ \delta x }-1 \right) }{ \delta x } \\ \\ ={ e }^{ x }\cdot 1\\ \\ ={ e }^{ x }\\ \\ Therefore,\quad if:\\ \\ y={ e }^{ x },\\ \\ \frac { dy }{ dx } ={ e }^{ x }

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