Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

Sec. 1] Calculating Derivatives Directly 43
calculate Ax and A#, corresponding to a change in the argument:
a) fromx=l to x=l.l;
b) from x=3 to x = 2.
Solution. We have
a) Ax=l. 1 1=0.1,
Ai/ = (l.l 2
5-1.1 + 6) (I 2
b) Ax = 2 3 = 1,
At/ = (2* 5-2-1-6) (3* 5-3 -f- 6)--=0.
5- 1+6) = 0.29;
Example 2. In the case of the hyperbola y = ,
find the slope
of the
secant passing through the points M ( 3, -- and N ) { 10, -r^ ) .
V '
J
1 1 7
Solution. Here, Ax=10 3 = 7 and Ay = ^ 4= 5*- Hence,
1U o 5U
, AJ/ 1
Ax~~ 30'
2. The derivative. The derivative y'=j- of a function y-=f(x) with re-
spect to the argument x is the limit of the ratio -r^ when Ax approaches zero;
that is.
y>= lim >.
AJC -> o A*
The magnitude of the derivative yields the slope of the tangent MT to the
graph of the function y = f(x) at the point x (Fig. 11):
y'
tan q>.
Finding the derivative /' is usually called differentiation of the function. The
derivative y'=f' (x) is the rate of change of the function at the point x.
Example 3. Find the derivative of the function
and
Hence,
and
y = x*.
Solution. From formula (1) we have
Ay = (*+ A*)*
5*. One-sided derivatives. The expressions
x i
'= lim L^ lim
Ax AJC->O
2*Ax+ (Ax) 1
/'_(*)= lim f (*+**)-/(*)
AJ:-*--O Ax
/(x)= lim
Ax