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

The point of contact is (1, 1.5403)

So, the equation of the tangent is

The tangent crosses the curve, elsewhere.

The normal to the curve is given by

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

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To find the equation of the tangent line at (x = 1), first find the derivative of the function (y = x + \cos(x)). Then, evaluate the derivative at (x = 1) to get the slope of the tangent line. Next, use the point-slope form of the equation of a line, where the point is ( (1, f(1)) ), and the slope is the derivative evaluated at (x = 1). This gives the equation of the tangent line. To find the equation of the normal line, calculate the negative reciprocal of the slope of the tangent line and use the same point to form the equation of the normal line.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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