Q.4. y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots \ldots \ldots\left(1+x^{2 n}\right), then the value of \frac{\mathrm{d} y}{\mathrm{~d} x} at x=0 is
A. 0
B. -1
C. 1
D. 2
Answer: C
Explanation
y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2 n}\right)Taking ‘log’ on both sides, we get
\log y=\log (1+x)+\log \left(1+x^{2}\right) +\log \left(1+x^{4}\right) +\ldots+\log \left(1+x^{2 n}\right)Differentiating w.r.t. x, we get
\frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\ldots+\frac{2 \mathrm{n} \times x^{2 \mathrm{n}-1}}{1+x^{2 \mathrm{n}}}At x=0, (\mathrm{i}) \Rightarrow y=1
\therefore \quad (\mathrm{ii}) \left.\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=0}=1+0+0+\ldots+0=1