How do you find the equation of the tangent and normal line to the curve $y=x+cosx$ at $x=1$ ?

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Answer 1 · 2016-12-26T03:03:14

$Tangent: y = 0.15852x+1.3818$ See Socratic graph for the curve, the point of contact and the tangent. Normal : $y = -6.3084x+7.8487$

$x = 1 radian = 57.296^o$

The point of contact is (1, 1.5403)

y' ar x = 1 is $m = 1-sin 57.2903^o=0.15852$ , nearly.

So, the equation of the tangent is

$y-1.5403=0.15852(x-1)$ Simplifying,

$y = 0.15852x+1.3818$ .

The tangent crosses the curve, elsewhere.

The normal to the curve is given by

$y-1.5403=-1/0.15852(x-1)$ , Simplifying,

$y = -6.3084x+7.8487$

graph{(y-0.16x-1.38)(y-x-cos x)(y+6.31x-7.84)=0 [-20, 20, -10, 10]}

How do you find the equation of the tangent and normal line to the curve y=x+cosxy=x+cosx at x=1x=1?