Explanation Given: The function, y = 2 x + 1 x + 2 and the point ( 1 , 1 ) 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 ) . 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 = 2 x + 1 x + 2 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 power rule and quotient rule of derivatives, the first derivative of the function is given by y ' = d d x ( 2 x + 1 x + 2 ) = ( x + 2 ) ⋅ d d x ( 2 x + 1 ) − ( 2 x + 1 ) ⋅ d d x ( x + 2 ) ( x

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Chapter 3.1, Problem 8E

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The equation of the tangent line to the graph of y=2x+1x+2 at the point (1,1) .

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Chapter 3.1, Problem 7E

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The equation of the tangent line to the graph of y = 2 x + 1 x + 2 at the point ( 1 , 1 ) .

Solution Summary: The author explains that the slope of the tangent line of a function f(x) is given by the first derivative at that point.

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The equation of the tangent line to the graph of y=2x+1x+2 at the point (1,1) .

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