Question

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)

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},f\left(x\right)=\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right) \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}, \\ \log f\left(x\right)=\log \left(1+x\right)+\log \left(1+x^2\right)+\log \left(1+x^4\right)+\log \left(1+x^8\right) \\ \Rightarrow \frac{d}{dx}\left\{\log f\left(x\right)\right\}=\frac{d}{dx}\left\{\log \left(1+x\right)+\log \left(1+x^2\right)+\log \left(1+x^4\right)+\log \left(1+x^8\right)\right\} \\ \Rightarrow \frac{1}{f\left(x\right)}f\text{'}\left(x\right)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8} \\ \Rightarrow f\text{'}\left(x\right)=\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right) \\ \Rightarrow f\text{'}\left(1\right)=\left\{1+\left(1\right)\right\}\left\{1+{\left(1\right)}^2\right\}\left\{1+{\left(1\right)}^4\right\}\left\{1+{\left(1\right)}^8\right\}\left\{\frac{1}{1+\left(1\right)}+\frac{2\left(1\right)}{1+{\left(1\right)}^2}+\frac{4{\left(1\right)}^3}{1+{\left(1\right)}^4}+\frac{8{\left(1\right)}^7}{1+{\left(1\right)}^8}\right\} \\ \Rightarrow f\text{'}\left(1\right)=2\times 2\times 2\times 2\left\{\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right\} \\ \Rightarrow f\text{'}\left(1\right)=2\times 2\times 2\times 2\times \frac{1}{2}\left\{1+2+4+8\right\} \\ \Rightarrow f\text{'}\left(1\right)=8\times 15=120 \end{gathered}$$