Differentiate the given function w.r.t. $x$ .
$y = e^{x} + e^{x^{2}} + . . . + e^{x^{5}}$

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Solution

$\frac{d}{d x} (e^{x} + e^{x^{2}} + . . . + e^{x^{5}})$
$= \frac{d}{d x} (e^{x}) + \frac{d}{d x} (e^{x^{2}}) + \frac{d}{d x} (e^{x^{3}}) + \frac{d}{d x} (e^{x^{4}}) + \frac{d}{d x} (e^{x^{5}})$
$= e^{x} + [e^{x^{2}} . \frac{d}{d x} (x^{2})] + [e^{x^{3}} . \frac{d}{d x} (x^{3})] + [e^{x^{4}} . \frac{d}{d x} (x^{4})] + [e^{x^{5}} . \frac{d}{d x} (x^{5})]$
$= e^{x} + (e^{x^{2}} \times 2 x) + (e^{x^{3}} \times 3 x^{2}) + (e^{x^{4}} \times 4 x^{3}) + (e^{x^{5}} \times 5 x^{4})$
$= e^{x} + 2 x e^{x^{2}} + 3 x^{2} e^{x^{3}} + 4 x^{3} e^{x^{4}} + 5 x^{4} e^{x^{5}}$

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