Let $y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}})$. Then $y' - y''$ at $x = - 1$ is equal to
Solution
$$
\begin{aligned}
& y=\frac{1-x^{32}}{1-x}=1+x+x^2+x^3+\ldots+x^{31} \\\\
& y^{\prime}=1+2 x+3 x^2+\ldots+31 x^{30} \\\\
& y^{\prime}(-1)=1-2+3-4+\ldots+31=16 \\\\
& y^{\prime \prime}(x)=2+6 x+12 x^2+\ldots+31.30 x^{29} \\\\
& y^{\prime \prime}(-1)=2-6+12 \ldots 31.30=480 \\\\
& y^{\prime \prime}(-1)-y^{\prime}(-1)=-496
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions
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