# Free Online Courses in Math - Derivatives

## Course Content

1. Definition of Derivative
2. Main Properties
3. Finding Derivatives of
• Elementary Functions:
• constant function
• polynomial function
• 1/x function
• x1/2 function
• power function
• exponential function
• logarithmic function
• trigonometric function
• inverse function
• cyclometric function (inverse trigonometric functions)
• Complex Functions
4. Table of Elementary Derivatives
5. Derivative - Other Important Rules - Calculus
6. Derivative - Examples
• Derivative - Examples 1-10
• Derivative - Examples 11-20
• Derivative - Examples 21-30
• Derivative - Examples 31-40
• Derivative - Examples 41-50
• Derivative - Examples 51-60
• Derivative - Examples 61-70

Apart from that form of learning, one can also practise differentiation using Free Online Calculator.

## Literature.

G. M. Fichtenholz – Integral and differential calculus, vol. 1, PWN Warsaw 1999 (originally in Russian)

## Theory. Definition.

Let's consider a function f(x) defined in a given range γ. Next move from the x = x0 point with the step Δx > 0 or Δx < 0 such that the new value of x = x0 + Δx is still within the range γ. Then the value of the f function changes from f(x0) to f(x0 + Δx). If the the limit of the expression

[f(x0 + Δx) - f(x0)]/Δx

when

Δx tends to 0

exists it is called the derivative of f function over variable x in the x0 point.

### Derivatives. Notation.

• dy/dx or df(x0)/dx

by Leibniz

• y' or f'(x0)

by Lagrange

• Dy or Df(x0)

by Cauchy.

## Main Properties of Derivatives

### Derivatives. Fermat's Theorem.

#### Lemma.

Let the function f(x) have in the point x0 its finite derivative. If this derivative is

f'(x0) > 0 (or f'(x0) < 0)

then for x values that are close enough to x0 on its right the function f(x) is

f(x) > f(x0) (or f(x) < f(x0))

and for x values that are close enough to x0 on its left the function f(x) is

f(x) < f(x0) (or f(x) > f(x0))

In other words, one can say that function f(x) in the point x0 grows (or decays).

#### Proof.

Let's consider the case when f'(x) > 0. From definition of derivative of f(x)

[f(x0 + Δx) - f(x0)]/Δx when Δx → 0

we can find such a surrounding of x0

(x0 - ε, x0 + ε)

where

[f(x) - f(x0)]/[x - x0] > 0

and x ≠ x0. Then we have

x0 < x < x0 + ε

such that

x - x0 > 0

Then

[f(x) - f(x0)]/[x - x0] > 0
gives
f(x) - f(x0) > 0

thus

f(x) > f(x0)

on the other hand when

x0 - ε < x < x0

and

x - x0 < 0

we get

f(x) - f(x0) < 0

thus

f(x) > f(x0)

what was to be demonstrated (Q. E. D. quod erat demonstrandum).

### Derivatives. Fermat's theorem.

Let a function f(x) that is defined in a range γ have its minimum or maximum in an inner point p of γ. If in that point both right and left derivative of f over x exist and are finite then

f'(p) = 0

#### Proof.

Let the function f have maximum in the point p. Assume that

f'(p) ≠ 0

Then if f'(p) > 0 and if x > p then f(x) > f(p) or if f'(p) < 0 and if x < p then f(x) > f(p). In both cases f(p) cannot be the greatest value of f function in the γ range.

### Derivatives. Darboux's Theorem.

#### Theorem.

If a function f(x) have in the range [a, b] the finite derivative then this derivative i. e. the function f'(x) takes at least once each value between f'(a) and f'(b).

#### Proof.

Let's take a number P from the range (f'(a), f'(b)) such that

f'(a) > P > f'(b)

We can define a new continuous function g(x)

g(x) = f(x) - Px

having, in the range [a, b] the derivative

g'(x) = f'(x) - P

Because

g'(a) = f'(a) - P > 0 and g'(b) = f'(b) - P < 0

and g(x) function is continuous, the second Weierstrass theorem implies that must be such a point p where g(x) function takes its maximum value and, in this case, this point is within the [a, b] range i. e. a < p < b. So from the Fermat's theorem we have

g'(p) = f'(p) - P = 0

and it gives

f'(p) = P

what was to be demonstrated (Q. E. D. quod erat demonstrandum).

### Derivatives. Rolle's Theorem.

#### Theorem.

Let's assume that a function f(x):

1. is defined and continuous in the range [a, b]
2. has the finite derivative (i. e. the function f'(x) exists) at least in the range (a, b)
3. at both ends of the range [a, b] the function f has equal values f(a) = f(b).
Then between a and b one can find such a point p
a < p < b

that

f'(p) = 0.

#### Proof.

The function f(x) is continuous in the range [a, b] and from the second theorem of Weierstrass has in the range [a, b] maximum M and minimum m. Let's consider two cases:

1. M = m. Then in the range [a, b] function f(x) has the constant value. Then f'(x) = 0 everywhere in [a, b] and a point p can be any point from (a, b).
2. M > m. We know that f(a) = f(b) and then at least one of M and m values is taken in the point p somewhere inside the range [a, b] and then from the Fermat's theorem f'(p) = 0.

what was to be demonstrated (Q. E. D. quod erat demonstrandum).

### Derivatives. Lagrange's Theorem.

#### Theorem.

Let's assume that:

1. a function f(x) is defined and continuous in the range [a, b]
2. it has the finite derivative (i. e. the function f'(x) exists) at least in the range (a, b)

then between a and b is a point p such that

a < p < b

for which the following formula is fulfilled

[f(b) - f(a)]/[b - a] = f'(p).

#### Proof.

Let's define in the range (a, b) a new function F(x)

F(x) = f(x) - f(a) - [f(b) - f(a)]/[b - a] (x - a)

The function fulfills all assumptions of the Rolle's theorem. Its derivative

F'(x) = f'(x) - [f(b) - f(a)]/[b - a]

is finite in the range (a, b). And at both ends of [a, b]

F(a) = F(b) = 0.

The Rolle's theorem implies that within the range (a, b) is a point p such that

F'(p) = 0.

Thus we have

f'(p) - [f(b) - f(a)]/[b - a] = 0.

and finally

f'(p) = [f(b) - f(a)]/[b - a]

what was to be demonstrated (Q. E. D. quod erat demonstrandum).

### Derivatives. Limit of Derivative.

Let's assume that a function f(x) is continuous in the range [x0, x0 + H] (where H > 0) and has the finite derivative f'(x) for x > x0. If a finite or infinite limit of

f'(x) when x → x0+0

exists and equals K then the right derivative in the point x0 (because the limit is in x0+0) has the same value.

#### An example.

Let's consider the function

f(x) = -1/2 x-2

with the derivative

f'(x) = 1/x3

the limit of f'(x) when

x → +0

equals

+∞

and the limit of f'(x) when

x → -0

equals

-∞

It means that the function f(x) has only one-sided derivatives i. e. the right derivative +∞ and the left derivative -∞.

### Derivatives. Cauchy's Theorem.

#### Theorem.

Let's assume that:

1. function f(x) and g(x) are continuous in the range [a, b]
2. they have the finite derivatives (i. e. the function f'(x) and g'(x) exist) at least in the range (a, b)
3. g'(x) ≠ 0 in the range (a, b)

then between a and b is a point p such that

a < p < b

for which the following formula is fulfilled

[f(b) - f(a)]/[g(b) - g(a)] = f'(p)/g'(p).

#### Proof.

From the Rolle's theorem g(b) ≠ g(a). Let's consider in the range (a, b) a new function F(x) such that

F(x) = f(x) - f(a) - [f(b) - f(a)]/[g(b) - g(a)] (g(x) - g(a))

The function fulfills all assumptions of the Rolle's theorem. The function F(x) is continuous in the range [a, b] because in this range are continuous both functions f(x) and g(x). In (a, b) exists the derivative F'(x) and equals

F'(x) = f'(x) - [f(b) - f(a)]/[g(b) - g(a)]g'(x).

And at both ends of [a, b]

F(a) = F(b) = 0.

The Rolle's theorem implies that within the range (a, b) is a point p such that

F'(p) = 0.

Thus we have

f'(p) - [f(b) - f(a)]/[g(b) - g(a)]g'(p) = 0

and finally

f'(p) = [f(b) - f(a)]/[g(b) - g(a)]g'(p)
f'(p)/g'(p) = [f(b) - f(a)]/[g(b) - g(a)]

what was to be demonstrated (Q. E. D. quod erat demonstrandum). This theorem is called also the generalized theorem of average value.

### Finding Derivatives of Elementary Functions.

1. Constant functions.
If y = const then always Δy = 0. It does not depend on the Δx value.

2. Polynomial functions.
If y = xn where n is a natural number. Let's define Δx then Δy is

y + Δy = (x + Δx)n = xn + nxn-1 Δx + n(n-1)/2 xn-2Δx2 + …

thus

Δy = nxn-1 Δx + n(n-1)/2 xn-2Δx2 + …

and

Δy/Δx = nxn-1 + n(n-1)/2 xn-2Δx + …

when Δx → 0

y' = nxn-1
3. 1/x Function
If y = 1/x then y + Δy is

1/[x + Δx]

which gives

Δy = 1/(x + Δx) - 1/x = -Δx/[x(x + Δx)]
Δy/Δx = -1/[x(x + Δx)]

and finally when Δx → 0

y' = -1/x2
4. x1/2 Function
If y = x1/2 (for x > 0) then

y + Δy = [x + Δx]1/2

then

Δy = [x + Δx]1/2 - x1/2 = Δx/[x1/2 + (x + Δx)1/2]

and finally

Δy/Δx = 1/[x1/2 + (x + Δx)1/2]

when Δx → 0

y' = 1/[2x1/2]
5. Power functions.
If y = xμ where μ is a real number. Let's assume x ≠ 0

Δy/Δx = [(x + Δx)μ - xμ]/Δx = xμ-1[(1 + Δx/x)μ - 1]/[Δx/x]

when Δx → 0

y' = μxμ-1

because when Δx/x → 0

[(1 + Δx/x)μ - 1]/[Δx/x] → μ
6. Exponential functions.
If y = ax where a > 0, -∞ < x < +∞

Δy/Δx = [ax + Δx - ax]/Δx = ax[aΔx - 1]/Δx

when Δx → 0

y' = axln a

when a = e

y' = ex
7. Logarithmic functions.
If y = loga x where 0 < a ≠ 1, 0 < x < +∞

Δy/Δx = [loga(x + Δx) - loga x]/Δx = 1/x loga(1 + Δx/x)/[Δx/x]

when Δx → 0

y' = loga e/x

In the case of natural logarithm

y' = 1/x.
8. Trigonometric functions.
Let's consider y = sin(x) then

Δy/Δx = [sin(x + Δx) - sin(x)]/Δx = sin(1/2Δx)/[1/2Δx] cos(x + 1/2 Δx)

when Δx → 0

y' = cos(x)

because when Δx → 0

sin(Δx)/Δx = 1
9. Inverse functions.
Let's consider y = f(x) that has its inverse function and has in the point x0 finite and different from 0 derivative f'(x0). Then also the derivative of its inverse function g(y) such that x = g(y) equals 1/f(x0).

Δx/Δy = 1/[Δy/Δx]

when Δx → 0

g'(y0) = 1/f'(x0)
10. Ciclometric functions.
Let's consider y = arcsin(x) (-1 < x < 1 and -π/2 < y < π/2). y is the inverse function of x = sin(y). The derivative of x equals

x' = cos(y)

it implies that also the derivative of y exists

y' = 1/cos(y) = 1/[1 - sin2(y)]1/2 = +1/[1 - x2]1/2

the sign is set as plus because cos(y) > 0.

### Finding Derivatives of Complex Functions.

Theorem.
Let's assume that:

1. function y = f(x) has in the point x0 the derivative y'= f'(x0)
2. the function z = g(y) has in the point y0 = f(x0) the derivative z' = g'(y0). The complex function z = g(f(x)) has also in the point f(x0) its derivative that equals
[g(f(x))]' = g'(y0) f'(x0) = g'y(f(x0)) f'(x0)

or shorter

[g(f(x))]' = z'y y'x

### Derivatives. The Rules of Taking Derivatives

#### Table of Elementary Derivatives

1.  y = const y' = 0
2.  y = x; y' = 1
3.  y = xμ; y' = μ xμ - 1
4.  y = ax; y' = axln a
5.  y = loga(x); y' = loga(e)/x
6.  y = sin(x); y' = cos(x)
7.  y = cos(x); y' = -sin(x)
8.  y = tg(x); y' = 1/cos2(x)
9.  y = ctg(x); y' = -1/sin2(x)
10.  y = arcsin(x); y' = 1/[1 - x2]1/2
11.  y = arccos(x); y' = -1/[1 - x2]1/2
12.  y = arctg(x); y' = 1/[1 + x2]

#### Derivatives. Other Important Rules. Calculus

Let a function u = f(x) has in the point x the derivative u'.

1. then the function z = const u has also its derivative in the point x

lim Δz/Δx = const lim Δu/Δx = const u'
z' = const u'
2. Let a function v = g(x) has in the point x the derivative v'. Then the function y = u ± v has also the derivative in that point that equals

y + Δy = (u + Δu) ± (v + Δv)
Δy = Δu ± Δv

and

Δy/Δx = Δu/Δx ± Δv/Δx

finally in the limit when Δx → 0

Δu/Δx ± Δv/Δx → u' ± v'
3. Let a function v = g(x) has in the point x the derivative v'. Then the function y = uv has also the derivative in that point that equals

y + Δy = (u + Δu)(v + Δv)
Δy = Δu + u Δv + ΔuΔv

and

Δy/Δx = Δu/Δx v + u Δv/Δx + Δu/Δx Δv

finally in the limit when Δx → 0 also Δv → 0

Δu/Δx v + u Δv/Δx → u'v + uv'
4. Let a function v = g(x) differ from 0 and has in the point x the derivative v'. Then the function y = u/v has also the derivative in that point that equals

y + Δy = (u + Δu)/(v + Δv)
Δy = [Δu v - u Δv]/[v(v + Δv)]

and

Δy/Δx = [Δu/Δx v - u Δv/Δx]/[v(v + Δv)]

finally in the limit when Δx → 0 also Δv → 0

[Δu/Δx v - u Δv/Δx]/[v(v + Δv)] → [u'v - uv']/v2

### Derivatives. Examples.

Presented-below examples only show the way of finding and presenting solutions.

1. ### Example 1.

y = 2 sin3(3/x)1/2

#### derivative

y' = 6 sin2(3/x)1/231/2(-1/2x-3/2) = -33/2x-3/2sin2((3/x)1/2) cos((3/x)1/2)
2. ### Example 2.

y = sin2(x)/cos7(x) - 2/(5 cos5(x))

#### derivative

y' = [2sin(x)cos(x) cos7(x) - 7 cos6(x)(-sin(x)) sin2(x)]/cos14(x) - [50 cos4(x)(-sin(x)]/(25 cos10(x)) =
[2sin(x)cos8(x) + 7 cos6(x)sin3(x)]/cos14(x) - 2 sin(x)/cos6(x) =
[2sin(x)cos2(x) + 7 sin3(x)]/cos8(x) - 2 sin(x)/cos6(x) =
[2sin(x)cos2(x) + 7 sin3(x) - 2 sin(x) cos2(x)]/cos8(x) =
7 sin3(x)/cos8(x)
3. ### Example 3.

y = [sin(x) + (x + 2x1/2)1/2]1/2

#### derivative

y' = 1/2 [sin(x) + (x + 2x1/2)1/2]-1/2 [cos(x) + 1/2(x + 2x1/2)-1/2(1 + x-1/2)]
4. ### Example 4.

y = [1 + tg(x + 1/x)]1/2

#### derivative

y' = 1/2 [1 + tg(x + 1/x)]-1/2 [1 + tg2(x + 1/x)] [1 - x-2]
5. ### Example 5.

y = [3 tg(3x) - tg3(3x)]/[1 - 3tg2(3x)]

#### derivative

[(3/cos2(3x)3 - 3 tg2(3x)/cos2(3x)3)(1 - 3tg2(3x)) - (3tg(3x) - tg3(3x))(-6tg(3x)/cos2(3x)3)]/[1 - 3 tg2(3x)]2 =
[9/cos2(3x)(1 - tg2(3x))(1 - 3tg2(3x)) - 9/cos2(3x)(3tg(3x) - tg3(3x))(-2tg(3x))]/[1 - 3 tg2(3x)]2 =
9/cos2(3x)[1 - 3tg2(3x) - tg2(3x) + 3tg4(3x) + 6tg2(3x) - 2tg4(3x)]/[1 - 3 tg2(3x)]2 =
9/cos2(3x)[1 + 2tg2(3x) + tg4(3x)]/[1 - 3 tg2(3x)]2 =
9/cos2(3x)[1 + tg2(3x)]2/[1 - 3 tg2(3x)]2
6. ### Example 6.

y = tg(x) - ctg(x) - 2x

#### derivative

y' = 1/cos2(x) + 1/sin2(x) - 2 = 1 + tg2(x) + 1 + ctg2(x) - 2 = tg2(x) + ctg2(x)
7. ### Example 7.

y = arctg(3x)

#### derivative

y' = 3/[1 + (3x)2]
8. ### Example 8.

y = 7 arctg(x/2)

#### derivative

y' = 7/2 1/[1 + (x/2)2]
9. ### Example 9.

y = arcsin(1 - x)

#### derivative

y' = 1/[1 - (1 - x)2]1/2(-1) = -1/[2x - x2]1/2
10. ### Example 10.

y = arccos(1 - x2)1/2

#### derivative

y' = -1/[1 - (1 - x2)]1/2(1/2)(1 - x2)-1/2(-2x) =
x/|x|(1 - x2)-1/2 = sgn(x)/(1 - x2)1/2

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