perplexus.info :: Calculus : Fractional Function Differentiation

Fractional Function Differentiation (Posted on 2025-01-09)

Given

     (1+2x)^1/2(1+4x)^1/4(1+6x)^1/6...(1+98x)^1/98(1+100x)^1/100
y = -----------------------------------------------------
     (1+3x)^1/3(1+5x)^1/5(1+7x)^1/7...(1+99x)^1/99(1+101x)^1/101

Find y' at x = 0.

No Solution Yet Submitted by Danish Ahmed Khan

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Solution

| Comment 1 of 2

First note if f'(a)=0 and g'(a)=0 then (f*g)'(a)=0 from the chain rule.

The first term of the numerator and denominator have the form

(1+nx)^(1/n)/(1+(n+1)x)^(1/(n+1)) (as do the second terms, third, etc.)

The derivative of this expression at x=0 is 0.

(Work not shown. I don't feel like typing it in.)

So the product of all these terms also has derivative 0 at x=0.

Posted by Jer on 2025-01-09 13:52:16

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