00:01
This question we got to differentiate this particular equation.
00:03
So there is a multiplication of two functions actually.
00:07
So we've got to use the product rule definitely.
00:09
So our d .y over dx is going to be the first term remains as it is.
00:15
We differentiate the second term, which is natural log of x squared plus one.
00:20
Plus the second term remains as it is, which is natural log of x square plus one times.
00:25
We differentiate the first term, which is x square plus one.
00:30
So over here we have x square plus 1 is as it is.
00:35
Then a differentiation of natural log, we've got to use the chain rule over here.
00:38
So differentiation of natural log is got to be 1 over x square plus 1...