`y = (2x)/(x^2 + 1), (1, 1)` Find equations of the tangent line and the normal to the given curve at the specified point.

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Chapter 3, 3.2 - Problem 34 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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The slope of a tangent line to a function at a point is actually equal to the derivative of the function at that point.

`y=(2x)/(x^2+1)`

`dy/dx=(((x^2+1)*d/dx(2x))-(2x)*(d/dx(x^2+1)))/(x^2+1)^2`

`dy/dx=((x^2+1)*2-2x*2x)/(x^2+1)^2`

`dy/dx=(2x^2+2-4x^2)/(x^2+1)^2`

`dy/dx=(2-2x^2)/(x^2+1)^2`

Slope (m) of the tangent line at (1,1) = dy/dx at (1,1)

                                                      = `((2-2(1^2))/((1^2+1)^2))`

                                                      = 0

Equation of the tangent line can be found by using point slope form

y-y_1 = m(x-x_1)

y-1= 0(x-1)

y-1=0 

y=1

The normal line is perpendicular to the tangent line

Therefore slope of the normal line = -1/m

Equation of the normal is

y-y_1=(-1/m)(x-x_1)

m(y-y_1)=-(x-x_1)

0(y-1)=-(x-1)

x-1=0

x=1

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