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secondary 3 | A Maths
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Need help with this question. Thank you so much:)
Date Posted: 3 years ago
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x/(1 + x)
= (x + 1 - 1)/(1 + x)
= (x + 1)/(1 + x) - 1/(1 + x)
= 1 - 1/(1 + x)
So x/(1 + x) ≡ 1 - 1/(1 + x)
(≡ stands for 'equivalent to' or 'identical to' , in geometry it is 'congruent to' .
It is also used in modular arithmetic eg. 15 ≡ 3 mod 4 )
So, d/dx x/(1 + x)
= d/dx (1 - 1/(1 + x))
= d/dx (1 - (1 + x)-¹)
= 0 - (-1)(1 + x)-²
= 1/(1 + x)²
= (x + 1 - 1)/(1 + x)
= (x + 1)/(1 + x) - 1/(1 + x)
= 1 - 1/(1 + x)
So x/(1 + x) ≡ 1 - 1/(1 + x)
(≡ stands for 'equivalent to' or 'identical to' , in geometry it is 'congruent to' .
It is also used in modular arithmetic eg. 15 ≡ 3 mod 4 )
So, d/dx x/(1 + x)
= d/dx (1 - 1/(1 + x))
= d/dx (1 - (1 + x)-¹)
= 0 - (-1)(1 + x)-²
= 1/(1 + x)²
Thank you so much:)
Thank you so much
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