# Free Online Courses in Math - Derivatives

## Course Content - Part 2

1. Analytic Methods. Finding Derivatives of
2. Table of Elementary Derivatives
3. Derivative - Other Important Rules - Calculus
4. 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)

### Analytic Methods. 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. Cyclometric 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.

### Analytic Methods. 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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