Calculator Calculus

Calculus Level 2

I f y = x x + 1 + x + 1 x t h e n d 2 y d x 2 a t x = 1 i s ? If\quad y=\frac { x }{ x+1 } \quad +\frac { x+1 }{ x\quad } \quad then\quad \frac { { d }^{ 2 }\quad y }{ d{ x }^{ 2 } } \quad at\quad x=1\quad is?


The answer is 1.75.

Hrisheek Yadav by Hrisheek Yadav , 20, India

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1 solution

Arjen Vreugdenhil
Oct 28, 2015

The easiest way is to rewrite the function as y = 1 x − 1 x + 1 . y = \frac1x - \frac1{x+1}. The first derivative is y ′ = − 1 x 2 + 1 ( x + 1 ) 2 . y' = -\frac1{x^2} + \frac1{(x+1)^2}. The second derivative is y ′ ′ = 2 x 3 − 2 ( x + 1 ) 3 . y'' = \frac2{x^3} - \frac2{(x+1)^3}. Substitute x = 1 x =1 to find y ′ ′ ( 1 ) = 2 1 − 2 8 = 1.75 . y''(1) = \frac21 - \frac28 = \boxed{1.75}.

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