Explanation Given that the function is t = 2 r e r s 2 − tan ( 2 r + s ) . Here, the dependent variable is t and the independent variables are r and s . Differentiate t = 2 r e r s 2 − tan ( 2 r + s ) partially with respect to r as follows. ∂ t ∂ r = ∂ ∂ r ( 2 r e r s 2 − tan ( 2 r + s ) ) = ∂ ∂ r ( 2 r e r s 2 ) + ∂ ∂ r ( − tan ( 2 r + s ) ) = 2 e r s 2 ∂ ∂ r ( r ) + 2 r ∂ ∂ r ( e r s 2 ) − sec 2 ( 2 r + s ) ∂ ∂ r ( 2 r + s ) ( use Chain and Product rule ) = 2 e r s 2 ( 1 ) + 2 r e r s 2 ∂ ∂ r ( r s 2 ) − sec 2 ( 2 r + s ) ( 2 ) That is, ∂ t ∂ r = 2 e r s 2 + 2 r e r s 2 ( s 2 ) − 2 sec 2 ( 2 r + s ) = 2 e r s 2 + 2 r s 2 e r s 2 − 2 sec 2 ( 2 r + s ) Thus, the partial derivative of t = 2 r e r s 2 − tan ( 2 r + s ) with respect to r is 2 e r s 2 + 2 r s 2 e r s 2 − 2 sec 2 ( 2 r + s ) . Differentiate t = 2 r e r s 2 − tan ( 2 r + s ) partially with respect to s as follows

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Chapter 29.3, Problem 24E

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The partial derivatives of the function, ∂t∂r and ∂t∂s.

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Chapter 29.3, Problem 23E

Chapter 29.3, Problem 25E

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Chapter 29 Solutions

Basic Technical Mathematics

Ch. 29.1 - Practice Exercise If f(x, y) = 4xy2 – 3x2y, find...Ch. 29.1 - Prob. 1E Ch. 29.1 - Prob. 2E Ch. 29.1 - Prob. 3E Ch. 29.1 - Prob. 4E Ch. 29.1 - Prob. 5E Ch. 29.1 - Prob. 6E Ch. 29.1 - Prob. 7E Ch. 29.1 - Prob. 8E Ch. 29.1 - Prob. 9E

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The partial derivatives of the function, ∂ t ∂ r and ∂ t ∂ s .

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The partial derivatives of the function, ∂t∂r and ∂t∂s.

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Chapter 29 Solutions