NDA 2021
Exam NDA
Year 2021
Subject Mathematics
Chapter Differentiation NDA 2021 · Mathematics · Differentiation
If y=(1+x) (1+x^2) (1+x⁴)(1+x⁸)(1+x^ 16 ) then what is (d y)/(d x)...
If
y = ( 1 + x ) y=(1+x) y = ( 1 + x ) \left(1+x^2\right)
( 1 + x 4 ) ( 1 + x 8 ) ( 1 + x 16 ) \left(1+x⁴\right)\left(1+x⁸\right)\left(1+x^{16}\right) ( 1 + x 4 ) ( 1 + x 8 ) ( 1 + x 16 ) then what is
d y d x \frac{d y}{d x} d x d y at
x = 0 x=0 x = 0 equal to?
Step-by-step solution
y = ( 1 + x ) ( 1 + x ) 2 ( 1 + x 4 ) ( 1 + x 8 ) ( 1 + x 16 ) ln y = ln ( 1 + x ) + ln ( 1 + x 2 ) + ln ( 1 + x 4 ) + ln ( 1 + x 8 ) + ln ( 1 + x 16 ) 1 y d y d x = 1 1 − x + 2 x 1 + x 2 + 4 x 3 1 + x 4 + 8 x 7 1 + x 8 + 16 x 15 1 + x 16 d y d x = y [ 1 1 + x + 2 x 1 + x 2 + 4 x 3 1 + x 4 + 8 x 7 1 + x 8 + 16 x 15 1 + x 16 ] \begin{aligned}
& y=(1+x)(1+x)^2\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right) \\
& \ln y=\ln (1+x)+\ln \left(1+x^2\right)+\ln \left(1+x^4\right)+\ln \left(1+x^8\right) \\
& +\ln \left(1+x^{16}\right) \\
& \frac{1}{y} \frac{d y}{d x}=\frac{1}{1-x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\frac{8 x^7}{1+x^8}+\frac{16 x^{15}}{1+x^{16}} \\
& \frac{d y}{d x}=y\left[\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\frac{8 x^7}{1+x^8}+\frac{16 x^{15}}{1+x^{16}}\right]
\end{aligned} y = ( 1 + x ) ( 1 + x ) 2 ( 1 + x 4 ) ( 1 + x 8 ) ( 1 + x 16 ) ln y = ln ( 1 + x ) + ln ( 1 + x 2 ) + ln ( 1 + x 4 ) + ln ( 1 + x 8 ) + ln ( 1 + x 16 ) y 1 d x d y = 1 − x 1 + 1 + x 2 2 x + 1 + x 4 4 x 3 + 1 + x 8 8 x 7 + 1 + x 16 16 x 15 d x d y = y [ 1 + x 1 + 1 + x 2 2 x + 1 + x 4 4 x 3 + 1 + x 8 8 x 7 + 1 + x 16 16 x 15 ]
Put
x = 0 d y d x = 1 [ 1 + 0 + 0 + 0 + 0 ] = 1 x=0\frac{d y}{d x}=1[1+0+0+0+0]=1 x = 0 d x d y = 1 [ 1 + 0 + 0 + 0 + 0 ] = 1
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