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  4. Let y (x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y prime-y prime prime at x=-1 is equal to :

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Q. Let $y (x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)$. Then $y^{\prime}-y^{\prime \prime}$ at $x=-1$ is equal to :

JEE MainJEE Main 2023Limits and Derivatives

Solution:

$ y=\frac{1-x^{32}}{1-x} \Rightarrow y-x y=1-x^{32} $
$ y^{\prime}-x y^{\prime}-y=-32 x^{31}$
$ y^{\prime \prime}-x y^{\prime \prime}-y^{\prime}-y^{\prime}=-(32)(31) x^{30} $
at $ x=-1 \Rightarrow y^{\prime}-y^{\prime \prime}=496$