Explanation Given Information: The provided function is y = ln 2 x + 1 and x = 0 . Calculation: Consider the function, y = ln 2 x + 1 Find the slope of tangent line to the function y = ln 2 x + 1 by determining the derivative. Rewrite the function as, y = ln ( 2 x + 1 ) 1 2 = 1 2 ln ( 2 x + 1 ) Then, the derivative of the functions is, d y d x = d d x ( 1 2 ln ( 2 x + 1 ) ) Constant multiple rule of derivative of function f ( x ) is f ' ( c x ) = c f ' ( x ) where c is constant. Apply the constant multiple rule, d y d x = 1 2 d d x ln ( 2 x + 1 ) The derivative of natural logarithm of a function is d d x ln u = 1 u d u d x . Now, apply the above formula, d y d x = 1 2 ⋅ 1 ( 2 x + 1 ) d d x ( 2 x + 1 ) = 1 2 ( 2 x + 1 ) ( d d x ( 2 x ) + d d x ( 1 ) ) Simplify the derivative by using the constant multiple rule, d y d x = 1 2 ( 2 x + 1 ) ( 2 d x d x + d d x ( 1 ) ) = 1 2 ( 2 x + 1 ) ( 2 ⋅ 1 + 0 ) = 2 2 ( 2 x + 1 ) = 1 2 x + 1 So, the slope of the tangent to the graph of function y = ln 2 x + 1 is 1 2 x + 1 . Now, the slope at the point ( x 1 , y 1 ) is evaluated by calculating the value of d y d x | ( x 1 , y 1 )

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 11.5, Problem 69E

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To calculate: The equation of the tangent to the graph of the function y=ln2x+1 at the point where x=0. Also, sketch the curve along with the tangent line.

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Chapter 11.5, Problem 68E

Chapter 11.5, Problem 70E

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The equation of the tangent to the graph of the function y = ln 2 x + 1 at the point where x = 0 . Also, sketch the curve along with the tangent line.

Solution Summary: The author calculates the equation of tangent to the graph of the function y=mathrmlnsqrt2x+1 and sketched the curve along with the

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To calculate: The equation of the tangent to the graph of the function y=ln2x+1 at the point where x=0. Also, sketch the curve along with the tangent line.

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Chapter 11 Solutions