Question

hi guys please answer this question ..which is the correct option ​

Answer 1

Question :–

▪︎ If

$\: \: { \bold{y =(1 + x)(1 + {x}^{2} )(1 + {x}^{4}) .......(1 + {x}^{ {2}^{n} } ) }} \: \:$

then find ${ \bold{ \dfrac{dy}{dx} \: \: at \: \: x = 0}} .$

ANSWER :–

▪︎ ${ { \boxed{\bold{ \frac{dy}{dx} =1}}}}$

EXPLANATION :–

GIVEN :–

▪︎ A function –

$\implies { \bold{y =(1 + x)(1 + {x}^{2} )(1 + {x}^{4}) .......(1 + {x}^{ {2}^{n} } ) }}$

TO FIND :–

$\\ \implies { \bold{ \dfrac{dy}{dx} \: \: at \: \: x = 0}}$

SOLUTION :–

▪︎ Given function is –

$\\ { \implies \bold{y =(1 + x)(1 + {x}^{2} )(1 + {x}^{4}) .......(1 + {x}^{ {2}^{n} } ) }}$

▪︎ Now take log on both sides –

$\\ { \implies \bold{ log(y) = log[(1 + x)(1 + {x}^{2} )(1 + {x}^{4}) .......(1 + {x}^{ {2}^{n} } )]}}$

▪︎ We know that –

$\\ { \implies \bold{ log(m.n) = log(m). log(n) }}$

▪︎ So that –

$\\ { \implies \bold{ log(y) = log(1 + x) + log(1 + {x}^{2} ) + log(1 + {x}^{4}) + ....... + log(1 + {x}^{ {2}^{n} } )}}$

▪︎Now Differentiate with respect to 'x' –

$\\ { \implies \bold{ \dfrac{1}{y}. \frac{dy}{dx} = \frac{1}{1 + x} + \frac{1}{1 + {x}^{2} } (2x)+ \frac{1}{1 + {x}^{4} } (4 {x}^{3} ) + ....... + \frac{1}{1 + {x}^{ {2}^{n} } }[ {2}^{n} ({x}^{ {2}^{n} - 1})] }}$

$\\ { \implies \bold{ \frac{dy}{dx} = y [ \frac{1}{1 + x} + \frac{1}{1 + {x}^{2} } (2x)+ \frac{1}{1 + {x}^{4} } (4 {x}^{3} ) + ....... + \frac{1}{1 + {x}^{ {2}^{n} } }[ {2}^{n} ({x}^{ {2}^{n} - 1})]]}}$

▪︎ Now put x = 0 –

$\\ { \implies \bold{ \frac{dy}{dx} = (y) [ \frac{1}{1 + (0)} + \frac{1}{1 + {(0)}^{2} } (2 \times 0)+ \frac{1}{1 + {(0)}^{4} } (4 {(0)}^{3} ) + ....... + \frac{1}{1 + {(0)}^{ {2}^{n} } }[ {2}^{n} ({(0)}^{ {2}^{n} - 1})]]}}$

$\\ { \implies \bold{ \frac{dy}{dx} = (y)[ \frac{1}{1} + \frac{1}{1 } \times 0+ \frac{1}{1 } \times 0+ ....... + \frac{1}{1} \times 0]}}$

▪︎Now put x = 0 in y –

$\\ \implies { \bold{y =(1 + 0)(1 + {0}^{2} )(1 + {0}^{4}) .......(1 + {0}^{ {2}^{n} } ) }}$

$\\ \implies { \bold{y =(1)(1)(1) .......(1) }}$

$\\ \implies { \bold{y = 1}}$

▪︎ Now put y = 1 –

$\\ { \implies \bold{ \frac{dy}{dx} = (1)[1 + 0]}}$

$\\ { \implies { \boxed{\bold{ \frac{dy}{dx} =1}}}}$