00:01
In this problem of differentiation, we have to estimate the value of 0 .98 square divided by 2 .01 q plus 1 during linear approximation.
00:33
So first, let's write this as f of xxema y.
00:48
We'll introduce simi such that.
00:51
So we replace 0 .98 with x.
00:56
So that is x square and 2 .01 with y, y cube plus 1.
01:09
So we have f of 0 .98 mark 2 .01, say where to 0 .98.
01:25
So now we just have to find here where a and b corresponds.
01:54
To the closest integer, say one command two.
02:00
Therefore, we have to find the value of f of 1 minus 0 .02, where the small change in x and 2 plus 0 .71.
02:25
So before proceeding, we will recollect the definition for linear approximation where if f of x comma y is differentiable at a comma b then it is said to be linearly linear or the linear approximation is given by the formula f of a plus delta x comma b plus delta y where delta x and delta y small change in x and y respectively is approximately equal to f of a comma b plus f x of a comma b into delta x plus f y of a comma b into delta x.
03:14
So first let's verify whether it's differentiable.
03:18
Now f of x comma y is equal to it's given by the equation x y by y cube plus 1 so f of a comma b here corresponds to 1 comma 2 that is b2 1 by 1 by 9 f x of x .m.
03:55
Y differentiating the function partially with respect to x will get 2x divided keeping y as constant so f1 of f x of f x of 1 .2 of f x of 1 .2 is going to 2 plus 2 divided by 8 plus 1 that is 2 by 9.
04:23
So f y of x and how y will be equal to x squared as constant 2.
04:49
So differentiating 1 by yq is 1 partially with respect to y 2 .2, yq plus 1, then just minus 1 into yq plus 1.
05:21
Minus 1 minus 1, that's minus 2 into 3 y 12, that is equal to minus 3 x12, y 12 divided by by by cube plus 1 of 2.
05:41
So fyo from 2 will give us minus 3 into 4 that is 12, divided by y cube, there is 8, 9, 9 ,000.
06:01
Square 8 1 .2.
06:19
This implies f of x comma y is differentiable at 1 comma 2...