Explanation Given: The function, y = x x + 1 and the point ( 4 , 0.4 ) Concept Used: Slope of the tangent line of a function f ( x ) at any point x = c is given by the first derivative at that point. That is the slope of tangent line at x = c is given by f ′ ( c ) . Slope of the normal line at any point of the curve is equal to the negative reciprocal of the slope of tangent line. That is the slope of tangent line at x = c is given by − 1 f ′ ( c ) . The point-slope form of the line passing through the point ( x 1 , y 1 ) and having slope m is given by y − y 1 = m ( x − x 1 ) . Power rule of derivatives, d d x ( x n ) = n x n − 1 The quotient rule of derivatives: d d x [ f ( x ) g ( x ) ] = g ( x ) ⋅ d d x f ( x ) − f ( x ) ⋅ d d x g ( x ) ( g ( x ) ) 2 Derivative of any constant C is zero, that is, d d x ( C ) = 0 Calculation: Consider the function, y = x x + 1 In order to find the equation of tangent line, first find the slope of tangent line by finding the first derivative at the specified point. So, using the quotient rule and power rule of derivatives, the first derivative of the function is given by y ' = d d x ( x x + 1 ) = ( x + 1 ) ⋅ d d x ( x ) − ( x ) ⋅ d d x ( x + 1 ) ( x + 1 ) 2 = ( x + 1 ) ⋅ ( 1 2 x ) − ( x ) ⋅ ( 1 + 0 ) ( x + 1 ) 2 = x + 1 2 x − x ( x + 1 ) 2 = x + 1 − 2 x 2 x ( x + 1 ) 2 = 1 − x 2 x ( x + 1 ) 2 Hence the slope of the tangent line at ( 4 , 0.4 ) is given by m = y ' | ( 4 , 0.4 ) = 1 − 4 2 4 ( 4 + 1 ) 2 = − 3 2 ⋅ 2 ( 5 ) 2 = − 3 4 ( 25 ) = − 3 100 So, now using the point-slope form, the equation of tangent line passing through the point ( x 1 , y 1 ) = ( 4 , 0

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Chapter 3.4, Problem 38E

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The equation of the tangent line and normal line to the graph of y=xx+1 at the point (4, 0.4) .

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The equation of the tangent line and normal line to the graph of y = x x + 1 at the point ( 4 , 0.4 ) .

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