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Question 29 of 35

Q.If q=1000+8p1−p2q=1000+8p_1-p_2, then ∂q∂p1\dfrac{\partial q}{\partial p_1} is :

(a) 10001000
(b) −1-1
(c) 1000−P21000-P_2
(d) 88
Puducherry TnboardTamil Nadu HSC First Year (DGE) Commerce Board 2024MCQ· 1mImportance★★★★★
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∂q∂p1=8\dfrac{\partial q}{\partial p_1}=8.

To find the partial derivative with respect to p1p_1, treat p2p_2 (and any constant) as fixed and differentiate term by term:

q=1000+8p1−p2.q=1000+8p_1-p_2.

  • ∂∂p1(1000)=0\dfrac{\partial}{\partial p_1}(1000)=0 (constant),
  • ∂∂p1(8p1)=8\dfrac{\partial}{\partial p_1}(8p_1)=8, …

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