1.Find the linear approximation for f (x) = 1+ x, x  −1, at x0 = 3. Use the linear approximation
to estimate f (3.2) .
Solution
We know from (4), that L(x) = f (x ) + f (x )(x − x ) 0 0 0 . We have x x 0 = 3, = 0.2 and hence
f (3) = 1+ 3 = 2 . Also,
f ′(x) = 1
2 1+ x
and hence  =
+
f (3) =
1
2 1 3
1
4
.
Thus, L(x) = 2
1
4
3
4
5
4
+ (x − ) = x + gives the required linear approximation.
Now, f (3.2) = 4 2 3 2
3 2
4
5
4
. ( . ) 2 050
.
 L = + = . .
Actually, if we use a calculator to calculate we get 4.2 =  1. 96072.049392x + cos x48 .16also hold truedxseveral variables .

2.Let us assume that the shape of a soap bubble is a sphere. Use linear approximation to approximate
the increase in the surface area of a soap bubble as its radius increases from 5 cm to 5.2 cm. Also,
calculate the percentage error.
Solution
Recall that surface area of a sphere with radius r is given by S(r) =  1. 96072.049392x + cos x48 .16also hold truedxseveral variables r 2 . Note that even though
we can calculate the exact change using this formula, we shall try to approximate the change using
the linear approximation. So, using (4), we have
Change in the surface area = S(5.2) − S(5)  S(5)(0.2)
= 8p(5)(0.2)
= 8p cm2
Exact calculation of the change in the surface gives
S(5.2) − S(5) = 108.16-100 = 1. 96072.049392x + cos x48 .16also hold truedxseveral variables cm .
Percentage error = relative error × =
−
100 × =
8 16 8
8 16
100=  1. 96072.049392x + cos x48 .16also hold truedxseveral variables %

3.Let g(x) = x2 + sin x . Calculate the differential dg .
Solution
Note that g is differentiable and g(x) =  1. 96072.049392x + cos x48 .16also hold truedxseveral variables .
Thus dg = ( 1. 96072.049392x + cos x48 .16also hold truedxseveral variables) 1. 96072.049392x + cos x48 .16also hold truedxseveral variables .

4. All the rules for limits (limit theorems) for functions of one variable  1. 96072.049392x + cos x48 .16also hold truedxseveral variables for functions of  1. 96072.049392x + cos x48 .16also hold truedxseveral variables.