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Find the derivative of the function given by $f(x) =( 1 + x )( 1 + x^2) ( 1 + x^4) ( 1 + x^8)$ and hence find $ f' (1).$

$\begin{array}{1 1} ( 1 + x )( 1 + x^2) ( 1 + x^4) ( 1 + x^8)\bigg(\large\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^{\Large 3}}{1-x^{\Large 4}}+\large\frac{8x^{\Large7}}{1-x^{\large 8}}\bigg) , f'(1)=60 \\ ( 1 + x )( 1 + x^2) ( 1 + x^4) ( 1 + x^8)\bigg(\large\frac{1}{1+x}+\frac{2x}{1-x^2}+\frac{4x^{\Large 3}}{1+x^{\Large 4}}+\large\frac{7x^{\Large7}}{1+x^{\large 8}}\bigg) , f'(1)=120\\ \bigg(\large\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^{\Large 3}}{1+x^{\Large 4}}+\large\frac{8x^{\Large7}}{1+x^{\large 8}}\bigg) , f'(1)=240 \\ ( 1 + x )( 1 + x^2) ( 1 + x^4) ( 1 + x^8)\bigg(\large\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^{\Large 3}}{1+x^{\Large 4}}+\large\frac{8x^{\Large7}}{1+x^{\large 8}}\bigg) , f'(1)=120\end{array} $

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Toolbox:
  • $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$
Step 1:
Let $y=( 1 + x )( 1 + x^2) ( 1 + x^4) ( 1 + x^8)$
Taking $\log$ on both sides
$\log y=\log( 1 + x )+\log( 1 + x^2) +\log( 1 + x^4)+\log ( 1 + x^8)$
Differentiating with respect to $x$
$\large\frac{1}{y}.\frac{dy}{dx}=\large\frac{1}{1+x}+\frac{1}{1+x^2}\frac{d}{dx}$$(1+x^2)+\large\frac{1}{1+x^4}\frac{d}{dx}$$(1+x^4)+\large\frac{1}{1+x^8}\frac{d}{dx}$$(1+x^8)$
$\qquad\;\;\;=\large\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^{\large 3}}{1+x^{\large 4}}+\large\frac{8x^7}{1+x^{\large 8}}$
$\large\frac{dy}{dx}=$$y\bigg(\large\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^{\large 3}}{1+x^{\large 4}}+\large\frac{8x^7}{1+x^{\large 8}}\bigg)$
$\quad\;=( 1 + x )( 1 + x^2) ( 1 + x^4) ( 1 + x^8)\bigg(\large\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^{\Large 3}}{1+x^{\Large 4}}+\large\frac{8x^{\Large7}}{1+x^{\large 8}}\bigg)$
Step 2:
At x=1
$f'(1)=(1+1)(1+1)(1+1)(1+1)\bigg(\large\frac{1}{1+1}+\large\frac{2(1)}{1+1}+\large\frac{4(1)}{1+1}+\large\frac{8(1)}{1+1}\bigg)$
$f'(1)=2.2.2.2\big(\large\frac{1}{2}+\large\frac{2}{2}+\large\frac{4}{2}+\large\frac{8}{2}\big)$
$\;\;\;\quad=16\big(\large\frac{1}{2}$$+1+2+4\big)$
$\;\;\;\quad=16\big(\large\frac{1+2+4+8}{2}\big)$
$\;\;\;\quad=16\big(\large\frac{15}{2}\big)$
$\;\;\;\quad=\large\frac{16\times 15}{2}$
$\;\;\;\quad=120$
answered May 9, 2013 by sreemathi.v
 

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