Free Online Courses in Math - Derivatives

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 + nx^n-1 Δx + n(n-1)/2 x^n-2Δx² + …

   thus

   Δy = nx^n-1 Δx + n(n-1)/2 x^n-2Δx² + …

   and

   Δy/Δx = nx^n-1 + n(n-1)/2 x^n-2Δx + …

   when Δx → 0

   y' = nx^n-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. x^1/2 Function
   If y = x^1/2 (for x > 0) then

   y + Δy = [x + Δx]^1/2

   then

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

   and finally

   Δy/Δx = 1/[x^1/2 + (x + Δx)^1/2]

   when Δx → 0

   y' = 1/[2x^1/2]

5. Power functions.
   If y = x^μ where μ is a real number. Let's assume x ≠ 0

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

   when Δx → 0

   y' = μx^μ-1

   because when Δx/x → 0

   [(1 + Δx/x)^μ - 1]/[Δx/x] → μ

6. Exponential functions.
   If y = a^x where a > 0, -∞ < x < +∞

   Δy/Δx = [a^{x + Δx} - a^x]/Δx = a^x[a^Δx - 1]/Δx

   when Δx → 0

   y' = a^xln a

   when a = e

   y' = e^x

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

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

   when Δx → 0

   y' = log_a 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)^1/23^1/2(-1/2x^-3/2) = -3^3/2x^-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 + 2x^1/2)^1/2]^1/2

   derivative

   y' = 1/2 [sin(x) + (x + 2x^1/2)^1/2]^-1/2 [cos(x) + 1/2(x + 2x^1/2)^-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