- Messages
- 112
- Reaction score
- 18
- Points
- 28
Oh ok.....so for example it was 2/y^3.........we could write this as 2ln(y^3)In order to apply logarithm on 1/y^3 the derivative of y^3 should be present in the numerator then we can use it.
Oh ok.....so for example it was 2/y^3.........we could write this as 2ln(y^3)In order to apply logarithm on 1/y^3 the derivative of y^3 should be present in the numerator then we can use it.
No ....derivative of y^3 is 3y^2 this should be present.Oh ok.....so for example it was 2/y^3.........we could write this as 2ln(y^3)
So 3y^2 should be the numerator?.......Always??No ....derivative of y^3 is 3y^2 this should be present.
Liek for all denominators....the numerator should be a derivative in order to use ln?So 3y^2 should be the numerator?.......Always??
yesSo 3y^2 should be the numerator?.......Always??
correctLiek for all denominators....the numerator should be a derivative in order to use ln?
Thanks <3correct
You are welcomeThanks <3
For remembrance it's given on the formula sheetThanks <3
Thanks.....once again!!!For remembrance it's given on the formula sheet
No problemThanks.....once again!!!
can anyone please send me the working of feb19 q3?
Just to simplify it more ..it's not a necessity .in the last step, why did you multiply and divide with square root 2 ?
thank u so muchhhhJust to simplify it more ..it's not a necessity .
And I just read the question it says no need to simplify the answer thus you can give the answer without multiplying and dividing by square root 2.
You are welcomethank u so muchhhh
I solved it rougly I hope it is understandableCan someone possibly show me the working of question 5, feb march paper 32 2019?