If A,B, and C are functions of time related by A^2 =B^3 C^4, and B(7)=3, C(7)=2, A'(7)=5, and C'(7)= -4, find B'(7)
Not sure how to do this question, can somebody help me please?
A^2 = B^3*C^4 ==> B^3 = (A/C^2)^2 ==> B = (A/C^2)^(2/3) ==>
B' = (2/3)[(A/C^2)^(-1/3)](A'C^2 - 2ACC')/C^4 ==>
B'(7)=(2/3)[(A(7)/C^2(7))^(-1/3)](A'(7)C^2(7) - 2A(7)C(7)C'(7))/C^4(7)
IWe know what C(7) and A'(7) is, so we only need A(7) and then substitute:
A^2(7) = B^3(7)*C^4(7) = (3^3)(2^4) = 27*16 = 432
Tonio