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### 1. Simplification of [tex]\(\frac{(\sec x - \tan x)(\sec x + \tan x)}{\tan x}\)[/tex]
First, we use the identity:
[tex]\[ (\sec x - \tan x)(\sec x + \tan x) = \sec^2 x - \tan^2 x \][/tex]
We know from the trigonometric identity:
[tex]\[ \sec^2 x - \tan^2 x = 1 \][/tex]
Thus, the expression simplifies to:
[tex]\[ \frac{1}{\tan x} = \cot x \][/tex]
So, the simplified form of [tex]\(\frac{(\sec x - \tan x)(\sec x + \tan x)}{\tan x}\)[/tex] is [tex]\(\cot x\)[/tex].
### 2. Differentiating [tex]\( y = -x^3 \)[/tex] from first principles
To differentiate [tex]\( y = -x^3 \)[/tex] from first principles, we'll use the definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \][/tex]
Calculating [tex]\( f(x+h) \)[/tex]:
[tex]\[ f(x+h) = -(x+h)^3 = -x^3 - 3x^2h - 3xh^2 - h^3 \][/tex]
Thus,
[tex]\[ f'(x) = \lim_{h \to 0} \frac{[-x^3 - 3x^2h - 3xh^2 - h^3] - (-x^3)}{h} \][/tex]
[tex]\[ = \lim_{h \to 0} \frac{-3x^2h - 3xh^2 - h^3}{h} \][/tex]
[tex]\[ = \lim_{h \to 0} [-3x^2 - 3xh - h^2] \][/tex]
Taking the limit as [tex]\( h \to 0 \)[/tex], we get:
[tex]\[ f'(x) = -3x^2 \][/tex]
Hence, the derivative of [tex]\( y = -x^3 \)[/tex] is [tex]\( \frac{dy}{dx} = -3x^2 \)[/tex].
### 3. Finding the turning points of [tex]\( y = x^3 - 12x^2 + 36x \)[/tex]
First, we find the first derivative of the function:
[tex]\[ y = x^3 - 12x^2 + 36x \][/tex]
Differentiating [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dy}{dx} = 3x^2 - 24x + 36 \][/tex]
To find the turning points, we set the first derivative equal to zero:
[tex]\[ 3x^2 - 24x + 36 = 0 \][/tex]
Solving the quadratic equation:
[tex]\[ x^2 - 8x + 12 = 0 \][/tex]
[tex]\[ (x - 2)(x - 6) = 0 \][/tex]
[tex]\[ x = 2 \quad \text{or} \quad x = 6 \][/tex]
We now find the coordinates of the turning points by substituting these [tex]\( x \)[/tex]-values back into the original function [tex]\( y \)[/tex]:
[tex]\[ \text{At } x = 2: \quad y = 2^3 - 12(2)^2 + 36(2) = 8 - 48 + 72 = 32 \][/tex]
[tex]\[ \text{At } x = 6: \quad y = 6^3 - 12(6)^2 + 36(6) = 216 - 432 + 216 = 0 \][/tex]
Thus, the turning points are at [tex]\( (2, 32) \)[/tex] and [tex]\( (6, 0) \)[/tex].
### 4. Differentiating [tex]\( y = 5\sqrt{\ln x} \)[/tex] using the chain rule
Given:
[tex]\[ y = 5\sqrt{\ln x} = 5(\ln x)^{1/2} \][/tex]
Let [tex]\( u = \ln x \)[/tex], so [tex]\( y = 5u^{1/2} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \][/tex]
First, find [tex]\( \frac{dy}{du} \)[/tex]:
[tex]\[ \frac{dy}{du} = 5 \cdot \frac{1}{2}u^{-1/2} = \frac{5}{2}u^{-1/2} = \frac{5}{2\sqrt{u}} \][/tex]
Next, find [tex]\( \frac{du}{dx} \)[/tex]:
[tex]\[ \frac{du}{dx} = \frac{d}{dx} (\ln x) = \frac{1}{x} \][/tex]
Combining these gives us:
[tex]\[ \frac{dy}{dx} = \frac{5}{2\sqrt{u}} \cdot \frac{1}{x} = \frac{5}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{5}{2x\sqrt{\ln x}} \][/tex]
Hence, the derivative of [tex]\( y = 5\sqrt{\ln x} \)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{5}{2x\sqrt{\ln x}} \][/tex]
### 1. Simplification of [tex]\(\frac{(\sec x - \tan x)(\sec x + \tan x)}{\tan x}\)[/tex]
First, we use the identity:
[tex]\[ (\sec x - \tan x)(\sec x + \tan x) = \sec^2 x - \tan^2 x \][/tex]
We know from the trigonometric identity:
[tex]\[ \sec^2 x - \tan^2 x = 1 \][/tex]
Thus, the expression simplifies to:
[tex]\[ \frac{1}{\tan x} = \cot x \][/tex]
So, the simplified form of [tex]\(\frac{(\sec x - \tan x)(\sec x + \tan x)}{\tan x}\)[/tex] is [tex]\(\cot x\)[/tex].
### 2. Differentiating [tex]\( y = -x^3 \)[/tex] from first principles
To differentiate [tex]\( y = -x^3 \)[/tex] from first principles, we'll use the definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \][/tex]
Calculating [tex]\( f(x+h) \)[/tex]:
[tex]\[ f(x+h) = -(x+h)^3 = -x^3 - 3x^2h - 3xh^2 - h^3 \][/tex]
Thus,
[tex]\[ f'(x) = \lim_{h \to 0} \frac{[-x^3 - 3x^2h - 3xh^2 - h^3] - (-x^3)}{h} \][/tex]
[tex]\[ = \lim_{h \to 0} \frac{-3x^2h - 3xh^2 - h^3}{h} \][/tex]
[tex]\[ = \lim_{h \to 0} [-3x^2 - 3xh - h^2] \][/tex]
Taking the limit as [tex]\( h \to 0 \)[/tex], we get:
[tex]\[ f'(x) = -3x^2 \][/tex]
Hence, the derivative of [tex]\( y = -x^3 \)[/tex] is [tex]\( \frac{dy}{dx} = -3x^2 \)[/tex].
### 3. Finding the turning points of [tex]\( y = x^3 - 12x^2 + 36x \)[/tex]
First, we find the first derivative of the function:
[tex]\[ y = x^3 - 12x^2 + 36x \][/tex]
Differentiating [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dy}{dx} = 3x^2 - 24x + 36 \][/tex]
To find the turning points, we set the first derivative equal to zero:
[tex]\[ 3x^2 - 24x + 36 = 0 \][/tex]
Solving the quadratic equation:
[tex]\[ x^2 - 8x + 12 = 0 \][/tex]
[tex]\[ (x - 2)(x - 6) = 0 \][/tex]
[tex]\[ x = 2 \quad \text{or} \quad x = 6 \][/tex]
We now find the coordinates of the turning points by substituting these [tex]\( x \)[/tex]-values back into the original function [tex]\( y \)[/tex]:
[tex]\[ \text{At } x = 2: \quad y = 2^3 - 12(2)^2 + 36(2) = 8 - 48 + 72 = 32 \][/tex]
[tex]\[ \text{At } x = 6: \quad y = 6^3 - 12(6)^2 + 36(6) = 216 - 432 + 216 = 0 \][/tex]
Thus, the turning points are at [tex]\( (2, 32) \)[/tex] and [tex]\( (6, 0) \)[/tex].
### 4. Differentiating [tex]\( y = 5\sqrt{\ln x} \)[/tex] using the chain rule
Given:
[tex]\[ y = 5\sqrt{\ln x} = 5(\ln x)^{1/2} \][/tex]
Let [tex]\( u = \ln x \)[/tex], so [tex]\( y = 5u^{1/2} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \][/tex]
First, find [tex]\( \frac{dy}{du} \)[/tex]:
[tex]\[ \frac{dy}{du} = 5 \cdot \frac{1}{2}u^{-1/2} = \frac{5}{2}u^{-1/2} = \frac{5}{2\sqrt{u}} \][/tex]
Next, find [tex]\( \frac{du}{dx} \)[/tex]:
[tex]\[ \frac{du}{dx} = \frac{d}{dx} (\ln x) = \frac{1}{x} \][/tex]
Combining these gives us:
[tex]\[ \frac{dy}{dx} = \frac{5}{2\sqrt{u}} \cdot \frac{1}{x} = \frac{5}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{5}{2x\sqrt{\ln x}} \][/tex]
Hence, the derivative of [tex]\( y = 5\sqrt{\ln x} \)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{5}{2x\sqrt{\ln x}} \][/tex]
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