Derivatives from First Principles

Differentiation - Exercise 2.1

• The exercise focuses on finding derivatives of functions using the definition of the derivative (first principles).
• The general approach involves finding the change in the function, dividing by the change in x, and then taking the limit as the change in x approaches zero.

Q.1 - Derivatives from First Principles

(i) f(x) = 2x² + 1

• Let y = 2x² + 1.
• Find y + δy = 2(x + δx)² + 1.
• Determine δy = 2(x + δx)² + 1 - y.
• Simplify δy = 2(x² + 2xδx + (δx)²) + 1 - (2x² + 1) = 4xδx + 2(δx)².
• Divide by δx to get δy/δx = 4x + 2δx.
• Taking the limit as δx → 0, dy/dx = 4x.

(ii) f(x) = 2 - √x

• Let y = 2 - √x.
• Find y + δy = 2 - √(x + δx).
• Determine δy = 2 - √(x + δx) - (2 - √x) = √x - √(x + δx).
• Apply the binomial expansion approach to rationalize the difference.
• After rationalization and simplification, divide by δx and take the limit as δx approaches zero.
• The final derivative is -1 / (2√x).

(iii) f(x) = 1/√x = x^(-1/2)

• Start by setting y = x^(-1/2).
• Calculate y + δy = (x + δx)^(-1/2).
• Express δy as (x + δx)^(-1/2) - x^(-1/2).
• Manipulate the equation using the binomial expansion.
• Divide δy by δx and find the limit as δx tends to zero, resulting in dy/dx = -1 / (2x^(3/2)).

(iv) f(x) = 1/x³ = x⁻³

• Let y = x⁻³.
• Find y + δy = (x + δx)⁻³.
• Determine δy = (x + δx)⁻³ - x⁻³.
• Use binomial expansion and dividing by δx while taking the limit as δx approaches zero to find dy/dx = -3/x⁴.

(v) f(x) = 1/(x - a) = (x - a)⁻¹

• Let y = (x - a)⁻¹.
• Find y + δy = (x + δx - a)⁻¹.
• Determine δy = (x + δx - a)⁻¹ - (x - a)⁻¹.
• Simplify using the binomial theorem.
• Divide by δx and take the limit as δx approaches 0, resulting in dy/dx = -1/(x - a)².

(vi) f(x) = x(x - 3) = x² - 3x

• Let y = x² - 3x.
• Find y + δy = (x + δx)² - 3(x + δx).
• Determine δy = (x + δx)² - 3(x + δx) - (x² - 3x).
• Simplify the expression.
• Divide by δx and take the limit as δx approaches 0, so dy/dx = 2x - 3.

(vii) f(x) = 2/x⁴ = 2x⁻⁴

• Let y = 2x⁻⁴.
• Find y + δy = 2(x + δx)⁻⁴.
• Determine δy = 2(x + δx)⁻⁴ - 2x⁻⁴.
• Use binomial expansion, divide by δx, take the limit to get dy/dx = -8/x⁵.

(viii) f(x) = x² + 1/x = x² + x⁻¹

• Let y = x² + x⁻¹.
• Find y + δy = (x + δx)² + (x + δx)⁻¹.
• Determine δy = (x + δx)² + (x + δx)⁻¹ - (x² + x⁻¹).
• Apply binomial approximation to the (x + δx)⁻¹ term.
• Divide by δx and take the limit to evaluate the derivative, dy/dx = 2x - 1/x².

(ix) f(x) = (x + 4)^(1/3)

• Let y = (x + 4)^(1/3).
• Then y + δy = (x + δx + 4)^(1/3).
• Thus δy = (x + δx + 4)^(1/3) - (x + 4)^(1/3).
• Simplify using the binomial expansion, divide by δx, and take the limit as δx → 0, resulting in dy/dx = 1 / [3(x + 4)^(2/3)].

(x) f(x) = x^(3/2)

• Let y = x^(3/2).
• Then y + δy = (x + δx)^(3/2).
• So, δy = (x + δx)^(3/2) - x^(3/2)
• Applying the manipulations like binomial series, dividing by δx, and applying limits, dy/dx = (3/2)√x.

(xi) f(x) = x^(5/2)

• Let y = x^(5/2)
• y + δy = (x + δx)^(5/2)
• Applying the definition, δy = (x + δx)^(5/2) - x^(5/2)
• Use binomial expansion as before, divide by δx and take the limit as δx tends to zero, the result dy/dx = (5/2)x^(3/2).

(xii) f(x) = x^m

• Let y = x^m
• y + δy = (x + δx)^m
• Then δy = (x + δx)^m - x^m
• Use binomial expansion, divide by δx, and take the limit to obtain dy/dx = mx^(m-1).

(xiii) f(x) = 1/x^m = x^(-m)

• Let y = x^(-m).
• Then y + δy = (x + δx)^(-m)
• Calculate δy = (x + δx)^(-m) - x^(-m).
• Applying binomial expansion, divide by δx, and take the limit as δx approaches zero, yielding dy/dx = -mx^(-m-1).

(xiv) f(x) = x^40

• Let y = x^40.
• Then y + δy = (x + δx)^40.
• Therefore, δy = (x + δx)^40 - x^40
• Use binomial expansion to simplify, divide by δx and finding the limit, result is dy/dx = 40x^39.

(xv) f(x) = x^(-100)

• Let y = x^(-100).
• y + δy = (x + δx)^(-100)
• δy = (x + δx)^(-100) - x^(-100).
• Simplify using binomials and taking limits leading to dy/dx = -100x^(-101).

Q.2 - Further Derivatives from First Principle

(i) f(x) = √(x + 2)

• Let y = √(x + 2) = (x + 2)^(1/2).
• Then y + δy = √(x + δx + 2) = (x + δx + 2)^(1/2).
• Apply binomial expansion to determine and simplify δy before dividing by δx and taking limits to obtain dy/dx = 1 / [2√(x + 2)].

(ii) f(x) = 1/√(x + a) = (x + a)^(-1/2)

• Let y = 1/√(x + a) = (x + a)^(-1/2).
• y + δy = 1/√(x + δx + a) = (x + δx + a)^(-1/2).
• Then δy = (x + δx + a)^(-1/2) - (x + a)^(-1/2).
• Expand using binomial, simplify, divide by δx, and take limit for the final derivative, dy/dx = -1 / [2(x + a)^(3/2)].