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Question

If y=1+x1!+x22!+x33!+, then dydx= ________

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Solution

We have,

y=1+x1!+x22!+x33!+......

y=1+x1!+x22!+x33!+......+xn1(n1)!+xnn!

On differentiating both sides w.r.t x, we get

dydx=0+1+2x2!+3x23!+......+(n1)xn2(n1)!+nxn1n!

dydx=1+x+x22+......+xn2(n2)!+xn1(n1)!

dydx=1+x+x22+......+xn2(n2)!+xn1(n1)!+xnn!xnn!

dydx=(1+x1!+x22+......+xn2(n2)!+xn1(n1)!+xnn!)xnn!

dydx=yxnn!

Hence, this is the answer.


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