Explanation Given Information: The function, f ( x , t ) = x 2 − t x 3 + t Formula used: For a function, f ( x , y ) = f ( y ) f ( x ) the partial derivative with respect to x is calculated as, ∂ ∂ x [ f ( x , y ) ] = ∂ ∂ x [ f ( y ) f ( x ) ] = [ f ( x ) ∂ ∂ x [ f ( y ) ] − f ( y ) ∂ ∂ x [ f ( x ) ] [ f ( x ) ] 2 ] Here, f ( x ) is the function of x and f ( y ) is the function of y . Calculation: Consider the following function, f ( x , t ) = x 2 − t x 3 + t For the calculation of f x , ∂ ∂ x [ f ( x , t ) ] = ∂ ∂ x [ x 2 − t x 3 + t ] = ( x 3 + t ) ∂ ∂ x ( x 2 − t ) − ( x 2 − t ) ∂ ∂ x ( x 3 + t ) ( x 3 + t ) 2 = ( x 3 + t ) ( 2 x + 0 ) − ( x 2 − t ) ( 3 x 3 − 1 − 0 ) ( x 3 + t ) 2 = ( x 3 + t ) ( 2 x ) − ( x 2 − t ) ( 2 x ) ( x 3 + t ) 2 Simplify it further, ∂ ∂ x [ f ( x , t ) ] = 2 x 4 + 2 x t 2 − 3 x 4 + 3 x 2 t ( x 3 + t ) 2 = − x 4 + 3 x 2 t + 2 x t ( x 3 + t ) 2 Since, f x = ∂ ∂ x [ f ( x , t ) ] Substitute − x 4 + 3 x 2 t + 2 x t ( x 3 + t ) 2 for ∂ ∂ x [ f ( x , t ) ]

Calculus and Its Applications (11th Edition)

11th Edition

ISBN: 9780321979391

Author: Marvin L. Bittinger, David J. Ellenbogen, Scott J. Surgent

Publisher: PEARSON

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Chapter 6.2, Problem 58E

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To calculate: The values of fx and ft for f(x,t)=x2−tx3+t.

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Chapter 6.2, Problem 57E

Chapter 6.2, Problem 59E

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Chapter 6 Solutions

Calculus and Its Applications (11th Edition)

Ch. 6.1 - Forf(x,y)=x23xy,find(0,2),f(2,3),andf(10,5).Ch. 6.1 - 2. . Ch. 6.1 - 3. . Ch. 6.1 - 4. . Ch. 6.1 - Forf(x,y)=Inx+y3,findf(e,2),f(e2,4),andf(e3,5).Ch. 6.1 - 6. . Ch. 6.1 - Forf(x,y,z)=x2y2+z2,findf(1,2,3)andf(2,1,3).Ch. 6.1 - 8. . Ch. 6.1 - In Exercises 9-14, determine the domain of each...Ch. 6.1 - In Exercises 9-14, determine the domain of each...

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