JEE Main 2023 — Differentiation Question with Solution

Given,

Function $y\left(x\right)=\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)$

Now multiplying and dividing by $\left(1-x\right)$ both side we get,

$\left(1-x\right)y\left(x\right)=\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)$

$\Rightarrow y\left(1-x\right)=1-x^{32}$ 

$\Rightarrow y-xy=1-x^{32} ......\left(1\right)$

Now differentiating equation $\left(1\right)$ on both sides with respect to $x$ we get,

$y^{\text{'}}-x{y}^{\text{'}}-y=-32x^{31}.......\left(2\right)$

Now at $x=-1$

$\Rightarrow y^{\text{'}}(-1)=16$

Now differentiating equation $\left(2\right)$ with respect to $x$ on both sides we get,

$\Rightarrow y^{"}-x{y}^{"}-y^{\text{'}}-y^{\text{'}}=-\left(32\right)\left(31\right)x^{30}$

Now at $x=-1$

$\Rightarrow y^{\text{'}\text{'}}(-1)=-480$

So, at $x=-1$, $y^{\text{'}}-y^{"}=16-(-480)=496$