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This is question from differentiation. V=KT 2/3 ---------------------------(i) dV=2KT -1/3 dT/3-----------------------------(ii) How we get equation (ii) from equation(i) Please please explain step by step,its my request.

This is question from differentiation.
V=KT2/3---------------------------(i)
dV=2KT-1/3dT/3-----------------------------(ii)
How we get equation (ii) from equation(i)
Please please explain step by step,its my request.

Grade:12th pass

2 Answers

Arun
25750 Points
5 years ago
Dear student
 
Have you read the chpater differentiation in mathematics.
 
In differentiation
 
Any function of y = xⁿ is diffrentiated with respect to x as
 
dy /dx = n xn-1 
 
Similarly in this case
 
DK/dT = (2/3) KT^-1/3
 
Now side transfer for dT
 
dK = (2/3) K T^-1/3. dT
 
Hope it helps
Kaushki
19 Points
5 years ago
No,I am a medical student.Thank you for explanation.

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