Correct Answer - Option 2 : 1
Concept:
- \(\rm \frac{d}{dx}x^{n} = nx^{n - 1}\)
- \(\rm \frac{d}{dx}ln \ x = \frac{1}{x}\)
- log m + lo n = log mn
Calculation:
y = (1 + x)(1 + x2)(1 + x4)(1 + x8)(1 + x16) -----(1)
At x = 0
y(0) = (1 + 0)(1 + 0)(1 + o)(1 + 0)(1 + 0)
⇒ y(0) = 1 -----(2)
Taking log on both side in equation (1)
ln y = ln[(1 + x)(1 + x2)(1 + x4)(1 + x8)(1 + x16)]
We know that,
log m + lo n = log mn
⇒ ln y = ln(1 + x) + ln(1 + x2) + ln(1 + x4) + ln(1 + x8) + ln(1 + x16)
Differentiating both side with respect to x,
⇒ \(\frac{1}{y}\frac{dy}{dx}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{16x^{15}}{1+x^{16}}\)
⇒ \(\frac{dy}{dx}=y[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{16x^{15}}{1+x^{16}}]\)
When x = 0, then
\(\frac{dy}{dx}=(1)(\frac{1}{1+0})\)
\(\therefore\ (\frac{dy}{dx})_{x=0}=1\)
