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प्रश्न
Find the derivative of the function f (x) given by \[f\left( x \right) = \left( 1 + x \right) \left( 1 + x^2 \right) \left( 1 + x^4 \right) \left( 1 + x^8 \right)\] and hence find `f' (1)` ?
उत्तर
\[\text{ We have }, f\left( x \right) = \left( 1 + x \right)\left( 1 + x^2 \right)\left( 1 + x^4 \right)\left( 1 + x^8 \right)\]
\[\text {Taking log on both sides }, \]
\[\log f\left( x \right) = \log\left( 1 + x \right) + \log\left( 1 + x^2 \right) + \log\left( 1 + x^4 \right) + \log\left( 1 + x^8 \right)\]
\[ \Rightarrow \frac{d}{dx}\left\{ \log f\left( x \right) \right\} = \frac{d}{dx}\left\{ \log\left( 1 + x \right) + \log\left( 1 + x^2 \right) + \log\left( 1 + x^4 \right) + \log\left( 1 + x^8 \right) \right\}\]
\[ \Rightarrow \frac{1}{f\left( x \right)}f'\left( x \right) = \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4 x^3}{1 + x^4} + \frac{8 x^7}{1 + x^8}\]
\[ \Rightarrow f'\left( x \right) = \left( 1 + x \right)\left( 1 + x^2 \right)\left( 1 + x^4 \right)\left( 1 + x^8 \right)\left( \frac{1}{1 + x} + \frac{2x}{1 + x^2} + \frac{4 x^3}{1 + x^4} + \frac{8 x^7}{1 + x^8} \right)\]
\[ \Rightarrow f'\left( 1 \right) = \left\{ 1 + \left( 1 \right) \right\}\left\{ 1 + \left( 1 \right)^2 \right\}\left\{ 1 + \left( 1 \right)^4 \right\}\left\{ 1 + \left( 1 \right)^8 \right\}\left\{ \frac{1}{1 + \left( 1 \right)} + \frac{2\left( 1 \right)}{1 + \left( 1 \right)^2} + \frac{4 \left( 1 \right)^3}{1 + \left( 1 \right)^4} + \frac{8 \left( 1 \right)^7}{1 + \left( 1 \right)^8} \right\}\]
\[ \Rightarrow f'\left( 1 \right) = 2 \times 2 \times 2 \times 2\left\{ \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} \right\}\]
\[ \Rightarrow f'\left( 1 \right) = 2 \times 2 \times 2 \times 2 \times \frac{1}{2}\left\{ 1 + 2 + 4 + 8 \right\}\]
\[ \Rightarrow f'\left( 1 \right) = 8 \times 15 = 120\]
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