Maths4

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Methods of dierentiation
4.1 Composition of functions

Let f and g be R R functions such that the range of g is contained in the domain of f . Then the mapping
x f (g(x))

is called the composition of the functions f and g . It is denoted by f g , that is
(f g)(x) = f (g(x)) .

For example, if f (x) = x and g(x) = 1 + x2 , then
(f g)(x) = 1 + x2 .



Attention, the composition of functions is not commutative!

In general f g = g f . If f and g are the functions from the above example, then
(g f )(x) = 1 + x ,

however this function is dened for x 0 only! Also, it may happen that f g is dened on an interval but g f is not dened at all. For example, if
g(x) = x , then (f g)(x) = -1 - x2 for x 0, but (g f )(x) = -1 - x4 is not dened for f (x) = -1 - x4

and

any real number. The theorem on the dierentiability of the composition of functions is a very strong tool for obtaining the derivative of more complicated functions.

Theorem 4.1 Suppose that g is dierentiable at x and f is dierentiable at g(x). Then f g is dierentiable at x and
(f g) (x) = f (g(x)) · g (x)

Introducing the notation k = g(x + h) - g(x) the dierence quotient for the function f g at x can be written in the following way:
f (g(x + h)) - f (g(x)) = h f (g(x) + k) - f (g(x)) g(x + h) - g(x) · k h

as long as g(x + h) - g(x) = 0. Then, by the continuity of g , k 0 as h 0, hence the limit of the expression on the right equals
f (g(x)) · g (x)

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4.2.

THE INVERSE FUNCTION

This argument fails if k = 0. In this case the proof is more complicated. We omit the details.

Example 4.1 Lete

F (x) = (1 + 3x - x2 )6 .

Then the derivative can be found without powering, if we recognize that with the notations f (x) = x6 and g(x) = 1 + 3x - x2 we have F = f g . Thus, according to the theorem:
F (x) = 6(1 + 3x - x2 )5 · (3 - 2x) .

4.2

The inverse function

Consider a function f : R R which is one-to-one on some interval. This means that f is either strictly increasing or strictly decreasing.

Denition 4.1 The inverse of f is the function f -1 whose domain is the range of f and whose range is the domain of f and for wich
(f -1 f )(x) = x

for all points of the domain of f . The inverse function can be determined so that we express y as a function of x from the equation
y = f (x) : x = f -1 (y) .

For instance, if f (x) = (2x + 5)3 , then it is clear that
f -1 (y) = 3 y-5 . 2

It is obvious that the graphs of f -1 and of f are symmetric with respect to the line y = x.

Theorem 4.2 Suppose that f is continuous, strictly monotonic on some interval, dierentiable at some interior point x of this interval and f (x) = 0. Then f -1 is also dierentiable at the point y = f (x) and
(f -1 ) (y0 ) = 1 . f (x0 )

We sketch the proof. Consider the dierence quotient:
f -1 (y + h) - f -1 (y) h

Let x and x+k points from the domain of f for which y = f (x) and y +h = f (x+k). Then the dierence quotient can be written as
x+k-x = f (x + k) - f (x) 1
f (x+k)-f (x) k

.

If h 0, then k 0 (this is not completely obvious!), hence the limit of the quotient on the right is 1/f (x). 2

4.2.

THE INVERSE FUNCTION

Example 4.2 Determine the derivative of the function
g(x) = n x

at some point x > 0. Observe that g is nothing else but the inverse of the power function f (x) = xn on the positive half-line, that is g(y) = f -1 (y). Therefore
g (y) =
1 1 1 1 = = · y n -1 n-1 f (x) nx n

as y = xn , hence
xn-1 = y

n-1 n

.

Using the argument of this example it is easy to see thatF (x) = xr is dierentiable for any rational exponent r at any point x > 0 and
F (x) = rxr-1 .

Example 4.3 Determine the derivative of the function
F (x) = 1 + x4 .

with the notations f (x) = x and g(x) = 1 + x4 we have F = f g. Therefore
4x3 F (x) = f (g(x)) · g (x) = 2 1 + x4

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4.2.

THE INVERSE FUNCTION

Exercise 4.1 Find the derivative of the following functions:

28. y = xx , 29. y = (sin x)cos x , 30. y = -2 arcsin 1 - x, 31. y = -2 arctan 32. y = ln (-x).
1-x x ,

1. y = sin x,
4



2. y = sin x4 , 3. y = sin 4x, 4. y = sin4 4x4 , 5. y = 6.
1-x , 1+x x y= , 1 - x2 1-x , 1+x 1 - sin x , 1 + sin x

7. y = ln 8. y = ln

9. y = ln sin x, 10. y = arcsin 2x, 11. y = arctan 12. y = ln lg x, 13. y = ln 10x , 14. y = x2x , 15. y = xsin x , 16. y = e-x ,
2

x+1 , x-1

17. y = -xe-x · e-e ,
-x

18. y =

sin x · arctan x2 , x 1 2 x 2 cos x . 2 sin2 x

19. y = ln tan - 20. y = x3 + 1 2 , 21. y =
x2 -2x x2 -4 ,

22. y = 2 sin x cos x, 23. y = sin2 x + cos2 x, 24. y = x2 + 2 sin x + 3, 25. y = xe + ex + xe ex , 26. y = 27. y =
tan x , 1 + tan2 x
4



ln3 ln6

5

x,

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4.3.

THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS

4.3

The exponential and logarithmic functions

Consider the exponential function of base e and its inverse, the logarithmic function of base e on the real line (the latter function is denoted by ln):
f (x) = ex f -1 (x) = ln x (x > 0) .

These are called exponential and logarithmic function of natural base. We would like to determine the derivatives of these functions. For this, we shall make use of the fact that x
x±

lim

1+

1 x

=e.

Determine rst the derivative of the logarithmic function of natural base at x0 = 1. Both the right-hand and the left-hand limit of the dierence quotient
ln(1 + h) - ln 1 = ln(1 + h)1/h h

equals ln e at 0 (assuming the continuity of the logarithmic function). Thus, the derivative at 0 is 1. The derivative of f (x) = ex at 0 can be determined by the rule of dierentiation of the inverse function:
f (0) = lim 1 eh - 1 = =1. h0 h (ln) (1)

From here we easily get the derivative of the exponential function at an arbitrary point x:
f (x) = lim eh - 1 ex+h - ex = ex · lim = ex h0 h0 h h

Using again the rule of dierentiation of the inverse function we easily get the derivative of the logarithmic function at an arbitrary x > 0:
(f -1 ) (x) = 1 eln x = 1 . x

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4.4.

HOMEWORK:

Example 4.4 As an example determine the derivative of the general power function
f (x) = x

at some point x > 0, where is an arbitrary real number. It is clear that
f (x) = x = e ln x

Hence, by the chain rule:
1 1 f (x) = e ln x = x = x-1 x x

This means that the derivative can be calculated in the same way as for rational powers.
4.4 Homework:

· Book ch6.8, ch6.10-11, ch7.5 · Practice the rules of direntiation (chain rule, exp and log functions) · Exercise given above + the exercise in ch6.8, ch6.10-11, ch7.5

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