Continuity and Differentiability — Class 12 Maths Solution

Let $f(x) = y$

$\Rightarrow$ $y = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})$

By taking log on both sides , we get

$\log y = \log [(1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})]$

$\Rightarrow$ $\log y = \log ((1 + x) + \log (1 + {x^2}) + \log (1 + {x^4}) + \log (1 + {x^8})$

Differentiating w.r.t. x, we get

$\cfrac{1}{y}\cfrac{{dy}}{{dx}} = \cfrac{1}{{(1 + x)}} + \cfrac{1}{{(1 + {x^2})}}(2x) + \cfrac{1}{{(1 + {x^4})}}(4{x^3}) + \cfrac{1}{{(1 + {x^8})}}(8{x^7})$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = y\left[ {\cfrac{1}{{1 + x}} + \cfrac{{2x}}{{1 + {x^2}}} + \cfrac{{4{x^3}}}{{(1 + {x^4})}} + \cfrac{{8{x^7}}}{{(1 + {x^8})}}} \right]$

therefore, $f'(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8}) \times \left[ {\cfrac{1}{{1 + x}} + \cfrac{{2x}}{{1 + {x^2}}} + \cfrac{{2{x^3}}}{{1 + {x^4}}} + \cfrac{{8{x^7}}}{{1 + {x^8}}}} \right]$

$\Rightarrow$ $f'(1) = (1 + 1)(1 + 1)(1 + 1)(1 + 1)$

$\left[ {\cfrac{1}{{1 + 1}} + \cfrac{{2(1)}}{{1 + 1}} + \cfrac{{4(1)}}{{1 + 1}} + \cfrac{{8(1)}}{{1 + 1}}} \right]$

$\Rightarrow$ $f'(1) = (2)(2)(2)(2)\left[ {\cfrac{1}{2} + \cfrac{2}{2} + \cfrac{4}{2} + \cfrac{8}{2}} \right]$

$\Rightarrow$ $f'(1) = 16\left\{ {\cfrac{{1 + 2 + 4 + 8}}{2}} \right\} = \cfrac{{16}}{2}(15) = 8(15) = 120$