Explanation Given information: The equation of curve is y = 1 1 + x 2 and the point is ( 1 , 1 2 ) . Formula used : The quotient rule of differentiation is, f ′ ( x ) = d d x [ f ( x ) g ( x ) ] = g ( x ) ⋅ d d x f ( x ) − f ( x ) ⋅ d d x g ( x ) [ g ( x ) ] 2 The above formula is defined when the denominator g ( x ) ≠ 0 . Calculation: Section1: The derivative of the given curve gives the slope of the tangent line so use the quotient rule of differentiation for the given function to get the derivative function f ′ ( x ) . f ′ ( x ) = d y d x = d d x [ 1 1 + x 2 ] = ( 1 + x 2 ) ⋅ d d x ( 1 ) − ( 1 ) ⋅ d d x ( 1 + x 2 ) ( 1 + x 2 ) 2 = ( 1 + x 2 ) ⋅ ( 0 ) − ( 1 ) ⋅ ( 0 + 2 x ) ( 1 + x 2 ) 2 Further solve the above equation for f ′ ( x ) . f ′ ( x ) = − 2 x ( 1 + x 2 ) 2 ( Apply distributive property ) (1) The derivative function gives the slope of line that is m = f ′ ( x ) and line is pass through given point ( 1 , 1 2 ) so substitute 1 for x in the equation (1) to get the value of slope. m = − 2 ( 1 ) ( 1 + 1 2 ) 2 m = − 2 ( 2 ) 2 m = − 2 4 m = − 1 2 The formula of point-slope form of the line is, y − y 1 = m ( x − x 1 ) (2) The tangent line has slope m = − 1 2 and passes through point ( 1 , 1 2 ) so substitute − 1 2 for m , 1 for x 1 and 1 2 for y 1 in the equation (2) to get the equation of the tangent line

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Author: Tan

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Chapter 3.2, Problem 51E

To determine

To find: The equation of the tangent line and the normal to the curve y=11+x2 at the given point (1,12) .

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Chapter 3.2, Problem 50E

Chapter 3.2, Problem 52E

Chapter 3 Solutions

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The equation of the tangent line and the normal to the curve y = 1 1 + x 2 at the given point ( 1 , 1 2 ) .

Solution Summary: The author explains the quotient rule of differentiation for the given function to get the derivative function.

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To find: The equation of the tangent line and the normal to the curve y=11+x2 at the given point (1,12) .

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Chapter 3 Solutions