If y- x x-1 - x-1 x- then d 2 y d x 2 at x-1 is equal to

The correct option is D 7 8 y= x x+1 + x+1 x ⇒ dy dx =( 1 ) ( x+1 ) x( 1 ) #160;( x+1 ) #160;^ 2 +( 1 ) ( x ) ( x+1 ) · 1 x ^ 2 ...

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Byju's Answer

Standard XII

Mathematics

Derivative from First Principle

If y= x x+1...

Question

If $y = \frac{x}{x + 1} + \frac{x + 1}{x}$ , then $\frac{d^{2} y}{d x^{2}}$ at $x = 1$ is equal to

\frac{7}{4}

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\frac{7}{8}

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\frac{1}{4}

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\frac{- 7}{8}

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\frac{- 7}{4}

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Solution

The correct option is D $\frac{- 7}{8}$
$y = \frac{x}{x + 1} + \frac{x + 1}{x}$
$\Rightarrow \frac{d y}{d x} = \frac{(1) (x + 1) - x (1)}{{(x + 1)}^{2}} + \frac{(1) (x) - (x + 1) \cdot 1}{x^{2}}$
$= \frac{1}{{(x + 1)}^{2}} - \frac{1}{x^{2}}$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- 2}{{(1 + x)}^{3}} + \frac{2}{x^{3}}$
On putting $x = 1$ , we get
${(\frac{d^{2} y}{d x^{2}})}_{x = 1} = \frac{- 2}{{(1 + 1)}^{3}} + \frac{2}{{(1)}^{3}}$
$= \frac{- 2}{8} + 2$
$= - \frac{1}{4} + 2 = \frac{7}{4}$

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If y=xx+1+x+1x, then d2ydx2 at x=1 is equal to

The correct option is D −78y=xx+1+x+1x⇒dydx=(1)(x+1)−x(1)(x+1)2+(1)(x)−(x+1)⋅1x2=1(x+1)2−1x2⇒d2ydx2=−2(1+x)3+2x3On putting x=1, we get(d2ydx2)x=1=−2(1+1)3+2(1)3=−28+2=−14+2=74

If $y = \frac{x}{x + 1} + \frac{x + 1}{x}$ , then $\frac{d^{2} y}{d x^{2}}$ at $x = 1$ is equal to