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*terrified me watching it one day before midsems o_o
relatable
Nptep lectures are very helpful
Anyone please confirm at 11:55 the value of partial derivative w.r.t. y is wrong. There should be sin only, not cos.
Yes u r right.
Very nice explanation🙏🙏🙏🙏, thanks a lot
Thank You very much sir, very well explained.
Sir problem 2 Partial derivative wrt y Here f(0, ∆y) doesn't exist because sin(1/x) at x =0 is not defined!
Yes bro
f(0, ∆y) does exist and it is equal to 0...see the definition of the function...f(x,y)=0 if x=0...and indeed here x=0...whatever the y may be.
Here (1/x) is bounded,so prefer to Y first,that's assume it zero and sir is right!
ii think here is some fault in printing and also sir's answer in problem 2
nice one
In problem fy(0, ∆y) does not exist why you say fy(0, ∆y) exist verify it.
maybe sin is bounded by [-1,1] and delta y^2 hence maybe it's 0?
*terrified me watching it one day before midsems o_o
relatable
Nptep lectures are very helpful
Anyone please confirm at 11:55 the value of partial derivative w.r.t. y is wrong. There should be sin only, not cos.
Yes u r right.
Very nice explanation🙏🙏🙏🙏, thanks a lot
Thank You very much sir, very well explained.
Sir problem 2
Partial derivative wrt y
Here f(0, ∆y) doesn't exist because sin(1/x) at x =0 is not defined!
Yes bro
f(0, ∆y) does exist and it is equal to 0...see the definition of the function...f(x,y)=0 if x=0...and indeed here x=0...whatever the y may be.
Here (1/x) is bounded,so prefer to Y first,that's assume it zero and sir is right!
ii think here is some fault in printing and also sir's answer in problem 2
nice one
In problem fy(0, ∆y) does not exist why you say fy(0, ∆y) exist verify it.
maybe sin is bounded by [-1,1] and delta y^2 hence maybe it's 0?