The eqation of the tangent to the curve y=84+x2 at x=2

dydx=−16x(4+x2)2,x=2dydx=−3264=−12,y=1Equation of the tangenty−1=−12(x−2)⇒x+2y−4=0

The eqation of the tangent to the curve y=84+x2 at x=2

dydx=−16x(4+x2)2,x=2dydx=−3264=−12,y=1Equation of the tangenty−1=−12(x−2)⇒x+2y−4=0

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The eqation of the tangent to the curve $y = \frac{8}{4 + x^{2}}$ at x=2

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2x+y+3=0

2x-y-3=0

x-2y+4=0

x+2y-4=0

answer is D.

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Detailed Solution

$\frac{d y}{d x} = \frac{- 16 x}{{(4 + x^{2})}^{2}}, x = 2$

$\frac{d y}{d x} = \frac{- 32}{64} = \frac{- 1}{2}, y = 1$

Equation of the tangent

$y - 1 = \frac{- 1}{2} (x - 2)$ $\Rightarrow x + 2 y - 4 = 0$

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The eqation of the tangent to the curve $y = \frac{8}{4 + x^{2}}$ at x=2