Question

If $y =\left(1+x\right)\left(1+x^{2}\right) ....\left(1+x^{100}\right),$ then $\frac{dx}{dy}$ at $x = 0$ is

• A. 100
• B. 0
• C. 1
• D. 5050

Answer 1

Answer: (C)

1

Answer 2

$y =\left(1+x\right)\left(1+x^{2}\right) ....\left(1+x^{100}\right),$
Differeptiating w.r.t.,,'y', we get
$1 = \frac{dx}{dy} \left[\left(1+x^{2}\right)\left(1+x^{3}\right)...\left(1+x^{100}\right)\right] + \left(1+x\right)\left(2x \frac{dx}{dy}\right)\left(\left(1+x^{3}\right)+...\left(1+x^{100}\right)\right)+......+\left[\left(1+x^{2}\right)\left(1+x^{3}\right)...\left(1+x^{99}\right)\right]\left(100x^{99} \frac{dx}{dy}\right)$
Now, $\frac{dx}{dy}$ at $x = 0$
$1 = \frac{dx}{dy} \left[\left(1+0\right)\left(1+0\right)....\left(1+0\right)+\left(1+0\right)\left(0\right)\left(1+0\right).....\left(1+0\right) \right] = \frac{dx}{dy}$
$\therefore \:\:\:\: \frac{dx}{dy} = 1$