Given y = e^x

From first principles we have dy/dx =`lim_(h->0)` (f(x+h) -f(x)) / h,

therefore for y = e^x , dy/dx = `lim_(h->0)` (e^(x+h) - e^x ) / h

= `lim_(h->0)` (e^x.e^h - e^x) / h

= e^x . `lim_(h->0)` (e^h - 1) / h

But we...

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Given y = e^x

From first principles we have dy/dx =`lim_(h->0)` (f(x+h) -f(x)) / h,

therefore for y = e^x , dy/dx = `lim_(h->0)` (e^(x+h) - e^x ) / h

= `lim_(h->0)` (e^x.e^h - e^x) / h

= e^x . `lim_(h->0)` (e^h - 1) / h

But we know that `lim_(h->0)` (e^t - 1) /t = 1, thus we get

dy/dx= e^x . 1 = e^x

thus dy/dx = e^x.

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