If y-1-x1-x21-x4-1-x2n- thendy-dx x-0-

Y=1 x^4n/1 x (dy/dx )_x=0=1 ...

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Formation of a Differential Equation from a General Solution

If y=1+x1+x21...

Question

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{2 n}), then {(\frac{d y}{d x})}_{x = 0} =$

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Similar questions

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . (1 + x^{2^{n}})

{(\frac{d y}{d x})}_{x = 0} =

y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots \dots (1 + x^{2^{n}})

\frac{d y}{d x}

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{2 n}), then {(\frac{d y}{d x})}_{x = 0} =$

y = (1 + x) (1 + x^{2}) (1 + x^{4}), t h e n \frac{d y}{d x} a t x = 0 i s

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{2^{n}})

\frac{d y}{d x}

x = 0

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