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

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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.

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.

Derivatives. Examples 1-10

  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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