Answer :
Certainly! Let's break down the problem step-by-step to calculate the required derivatives.
### 1.1 Given [tex]\( f(x) = 1 - 5x^2 \)[/tex], determine [tex]\( f'(x) \)[/tex] from first principles.
To find the derivative of a function [tex]\( f(x) \)[/tex] from first principles, we use the definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \][/tex]
For [tex]\( f(x) = 1 - 5x^2 \)[/tex]:
1. Calculate [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x + h) = 1 - 5(x+h)^2 \][/tex]
[tex]\[ = 1 - 5(x^2 + 2xh + h^2) \][/tex]
[tex]\[ = 1 - 5x^2 - 10xh - 5h^2 \][/tex]
2. Calculate [tex]\( f(x + h) - f(x) \)[/tex]:
[tex]\[ f(x + h) - f(x) = (1 - 5x^2 - 10xh - 5h^2) - (1 - 5x^2) \][/tex]
[tex]\[ = -10xh - 5h^2 \][/tex]
3. Substitute into the definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{-10xh - 5h^2}{h} \][/tex]
[tex]\[ = \lim_{h \to 0} (-10x - 5h) \][/tex]
When [tex]\( h \)[/tex] approaches 0, the term [tex]\( -5h \)[/tex] approaches 0, so:
[tex]\[ f'(x) = -10x \][/tex]
Therefore, [tex]\( f'(x) = -10x \)[/tex].
### 1.2 Differentiate with respect to [tex]\( x \)[/tex]:
#### 1.2.1 [tex]\( y = (2x - 1)^2 \)[/tex]
To differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex], we use the chain rule:
[tex]\[ y = (2x - 1)^2 \][/tex]
Differentiate the outer function first, then the inner function:
[tex]\[ \frac{dy}{dx} = 2(2x - 1) \cdot \frac{d}{dx}(2x - 1) \][/tex]
[tex]\[ = 2(2x - 1) \cdot 2 \][/tex]
[tex]\[ = 4(2x - 1) \][/tex]
[tex]\[ = 8x - 4 \][/tex]
Therefore, [tex]\( \frac{dy}{dx} = 8x - 4 \)[/tex].
#### 1.2.2 [tex]\( y = \sqrt{x} (1 - \sqrt[3]{x}) \)[/tex]
To differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex], apply the product rule combined with the chain rule. Let [tex]\( u = \sqrt{x} \)[/tex] and [tex]\( v = 1 - \sqrt[3]{x} \)[/tex].
[tex]\[ y = u \cdot v \][/tex]
First, find [tex]\( \frac{du}{dx} \)[/tex] and [tex]\( \frac{dv}{dx} \)[/tex]:
[tex]\[ u = x^{1/2} \Rightarrow \frac{du}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \][/tex]
[tex]\[ v = 1 - x^{1/3} \Rightarrow \frac{dv}{dx} = -\frac{1}{3} x^{-2/3} \][/tex]
Now apply the product rule:
[tex]\[ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = \sqrt{x} \left(-\frac{1}{3} x^{-2/3}\right) + (1 - x^{1/3}) \left(\frac{1}{2\sqrt{x}}\right) \][/tex]
Simplify each term:
[tex]\[ \frac{dy}{dx} = -\frac{1}{3} x^{1/2 - 2/3} + \frac{1 - x^{1/3}}{2x^{1/2}} \][/tex]
[tex]\[ = -\frac{1}{3} x^{-1/6} + \frac{1 - x^{1/3}}{2\sqrt{x}} \][/tex]
Therefore:
[tex]\[ \frac{dy}{dx} = -0.333333333333333/x^{0.166666666666667} + \frac{1 - x^{0.333333333333333}}{2\sqrt{x}} \][/tex]
#### 1.2.3 [tex]\( D_x \left[ \frac{8x^3 - 27}{2x - 3} \right] \)[/tex]
To differentiate [tex]\( \frac{8x^3 - 27}{2x - 3} \)[/tex] with respect to [tex]\( x \)[/tex], we use the quotient rule:
Let [tex]\( u = 8x^3 - 27 \)[/tex] and [tex]\( v = 2x - 3 \)[/tex].
Quotient rule states:
[tex]\[ \frac{du}{dx} = 24x^2 \][/tex]
[tex]\[ \frac{dv}{dx} = 2 \][/tex]
[tex]\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \][/tex]
Substitute [tex]\( u \)[/tex], [tex]\( v \)[/tex], [tex]\( \frac{du}{dx} \)[/tex], and [tex]\( \frac{dv}{dx} \)[/tex]:
[tex]\[ \frac{d}{dx} \left( \frac{8x^3 - 27}{2x - 3} \right) = \frac{(2x - 3) \cdot 24x^2 - (8x^3 - 27) \cdot 2}{(2x - 3)^2} \][/tex]
Simplify:
[tex]\[ = \frac{48x^2 (2x - 3) - 16x^3 + 54}{(2x - 3)^2} \][/tex]
[tex]\[ = \frac{96x^3 - 144x^2 - 16x^3 + 54}{(2x - 3)^2} \][/tex]
[tex]\[ = \frac{80x^3 - 144x^2 + 54}{(2x - 3)^2} \][/tex]
This comprehensively solves the differentiation tasks.
### 1.1 Given [tex]\( f(x) = 1 - 5x^2 \)[/tex], determine [tex]\( f'(x) \)[/tex] from first principles.
To find the derivative of a function [tex]\( f(x) \)[/tex] from first principles, we use the definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \][/tex]
For [tex]\( f(x) = 1 - 5x^2 \)[/tex]:
1. Calculate [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x + h) = 1 - 5(x+h)^2 \][/tex]
[tex]\[ = 1 - 5(x^2 + 2xh + h^2) \][/tex]
[tex]\[ = 1 - 5x^2 - 10xh - 5h^2 \][/tex]
2. Calculate [tex]\( f(x + h) - f(x) \)[/tex]:
[tex]\[ f(x + h) - f(x) = (1 - 5x^2 - 10xh - 5h^2) - (1 - 5x^2) \][/tex]
[tex]\[ = -10xh - 5h^2 \][/tex]
3. Substitute into the definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{-10xh - 5h^2}{h} \][/tex]
[tex]\[ = \lim_{h \to 0} (-10x - 5h) \][/tex]
When [tex]\( h \)[/tex] approaches 0, the term [tex]\( -5h \)[/tex] approaches 0, so:
[tex]\[ f'(x) = -10x \][/tex]
Therefore, [tex]\( f'(x) = -10x \)[/tex].
### 1.2 Differentiate with respect to [tex]\( x \)[/tex]:
#### 1.2.1 [tex]\( y = (2x - 1)^2 \)[/tex]
To differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex], we use the chain rule:
[tex]\[ y = (2x - 1)^2 \][/tex]
Differentiate the outer function first, then the inner function:
[tex]\[ \frac{dy}{dx} = 2(2x - 1) \cdot \frac{d}{dx}(2x - 1) \][/tex]
[tex]\[ = 2(2x - 1) \cdot 2 \][/tex]
[tex]\[ = 4(2x - 1) \][/tex]
[tex]\[ = 8x - 4 \][/tex]
Therefore, [tex]\( \frac{dy}{dx} = 8x - 4 \)[/tex].
#### 1.2.2 [tex]\( y = \sqrt{x} (1 - \sqrt[3]{x}) \)[/tex]
To differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex], apply the product rule combined with the chain rule. Let [tex]\( u = \sqrt{x} \)[/tex] and [tex]\( v = 1 - \sqrt[3]{x} \)[/tex].
[tex]\[ y = u \cdot v \][/tex]
First, find [tex]\( \frac{du}{dx} \)[/tex] and [tex]\( \frac{dv}{dx} \)[/tex]:
[tex]\[ u = x^{1/2} \Rightarrow \frac{du}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \][/tex]
[tex]\[ v = 1 - x^{1/3} \Rightarrow \frac{dv}{dx} = -\frac{1}{3} x^{-2/3} \][/tex]
Now apply the product rule:
[tex]\[ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = \sqrt{x} \left(-\frac{1}{3} x^{-2/3}\right) + (1 - x^{1/3}) \left(\frac{1}{2\sqrt{x}}\right) \][/tex]
Simplify each term:
[tex]\[ \frac{dy}{dx} = -\frac{1}{3} x^{1/2 - 2/3} + \frac{1 - x^{1/3}}{2x^{1/2}} \][/tex]
[tex]\[ = -\frac{1}{3} x^{-1/6} + \frac{1 - x^{1/3}}{2\sqrt{x}} \][/tex]
Therefore:
[tex]\[ \frac{dy}{dx} = -0.333333333333333/x^{0.166666666666667} + \frac{1 - x^{0.333333333333333}}{2\sqrt{x}} \][/tex]
#### 1.2.3 [tex]\( D_x \left[ \frac{8x^3 - 27}{2x - 3} \right] \)[/tex]
To differentiate [tex]\( \frac{8x^3 - 27}{2x - 3} \)[/tex] with respect to [tex]\( x \)[/tex], we use the quotient rule:
Let [tex]\( u = 8x^3 - 27 \)[/tex] and [tex]\( v = 2x - 3 \)[/tex].
Quotient rule states:
[tex]\[ \frac{du}{dx} = 24x^2 \][/tex]
[tex]\[ \frac{dv}{dx} = 2 \][/tex]
[tex]\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \][/tex]
Substitute [tex]\( u \)[/tex], [tex]\( v \)[/tex], [tex]\( \frac{du}{dx} \)[/tex], and [tex]\( \frac{dv}{dx} \)[/tex]:
[tex]\[ \frac{d}{dx} \left( \frac{8x^3 - 27}{2x - 3} \right) = \frac{(2x - 3) \cdot 24x^2 - (8x^3 - 27) \cdot 2}{(2x - 3)^2} \][/tex]
Simplify:
[tex]\[ = \frac{48x^2 (2x - 3) - 16x^3 + 54}{(2x - 3)^2} \][/tex]
[tex]\[ = \frac{96x^3 - 144x^2 - 16x^3 + 54}{(2x - 3)^2} \][/tex]
[tex]\[ = \frac{80x^3 - 144x^2 + 54}{(2x - 3)^2} \][/tex]
This comprehensively solves the differentiation tasks.
