Explanation Given Information: The provided function is y = ln ( 3 x + 2 x 2 − 5 ) 1 / 4 . Formula Used: The quotient rule for any positive numbers M and N , ln M N = ln M − ln N . The power rule for any positive number M is such that, ln M p = p ⋅ ln M , where p is any real number. Coefficient rule for a constant c is such that, if f ( x ) = c ⋅ u ( x ) , where u ( x ) is a differentiable function of x , then f ′ ( x ) = c ⋅ u ′ ( x ) . Derivative of natural logarithmic functions is such that, if y = ln u , where u is a differentiable function of x then d y d x = 1 u ⋅ d u d x . Power of x rule for a real number n is such that, if f ( x ) = x n then f ′ ( x ) = n x n − 1 . Constant function rule for a constant c is such that, if f ( x ) = c then f ′ ( x ) = 0 . Calculation: Consider the function, y = ln ( 3 x + 2 x 2 − 5 ) 1 / 4 Use power rule for logarithms, y = 1 4 ln ( 3 x + 2 x 2 − 5 ) Use quotient rule for logarithms, y = 1 4 [ ln ( 3 x + 2 ) − ln ( x 2 − 5 ) ] Differentiate with respect to x , d y d x = d d x { 1 4 [ ln ( 3 x + 2 ) − ln ( x 2 − 5 ) ] } = 1 4 { d d x [ ln ( 3 x +

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Author: HARSHBARGER

Publisher: CENGAGE L

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Chapter 11.1, Problem 22E

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To calculate: The value of dydx for the function y=ln(3x+2x2−5)1/4.

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Chapter 11.1, Problem 21E

Chapter 11.1, Problem 23E

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Chapter 11 Solutions

MATHEMATICAL APPLICATIONS FOR THE MGT

Ch. 11.1 - 1. Ch. 11.1 - 2. If Ch. 11.1 - Prob. 3CP Ch. 11.1 - 4. Find . Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Find the derivatives of the functions in Problems...

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The value of d y d x for the function y = ln ( 3 x + 2 x 2 − 5 ) 1 / 4 .

Solution Summary: The author explains the function y=mathrmlnleft and the power rule for any positive number M.

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To calculate: The value of dydx for the function y=ln(3x+2x2−5)1/4.

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Chapter 11 Solutions