NSC Mathematics Calculus: 8.1 Bepaal $f'(x)$ vanuit eerste

8.1 Bepaal $f'(x)$ vanuit eerste beginsels indien dit gegee word dat $f(x) = x^2 - 5$ - NSC Mathematics - Question 8 - 2018 - Paper 1

Question 8

8.1 Bepaal $f'(x)$ vanuit eerste beginsels indien dit gegee word dat $f(x) = x^2 - 5$. 8.2 Bepaal $rac{dy}{dx}$ indien: 8.2.1 $y = 3x^3 + 6x^2 + x - 4$ 8.2.2 $yx... show full transcript

Worked Solution & Example Answer:8.1 Bepaal $f'(x)$ vanuit eerste beginsels indien dit gegee word dat $f(x) = x^2 - 5$ - NSC Mathematics - Question 8 - 2018 - Paper 1

Step 1

Bepaal $f'(x)$ vanuit eerste beginsels indien dit gegee word dat $f(x) = x^2 - 5$

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Answer

To determine the derivative from first principles, we start with the definition:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Given that $f(x) = x^2 - 5$ :

Calculate $f(x+h)$ : $f(x+h) = (x+h)^2 - 5 = x^2 + 2xh + h^2 - 5$
Subtract $f(x)$ from $f(x+h)$ : $f(x+h) - f(x) = (x^2 + 2xh + h^2 - 5) - (x^2 - 5) = 2xh + h^2$
Substitute this back into the limit: $f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}$
Simplify: $= \lim_{h \to 0} (2x + h)$ $= 2x$

Step 2

Bepaal $\frac{dy}{dx}$ indien: $y = 3x^3 + 6x^2 + x - 4$

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Answer

To find $rac{dy}{dx}$ , we differentiate $y$ with respect to $x$ :

Differentiate each term:
- The derivative of $3x^3$ is $9x^2$ .
- The derivative of $6x^2$ is $12x$ .
- The derivative of $x$ is $1$ .
- The derivative of $-4$ is $0$ .
Combine the derivatives: $\frac{dy}{dx} = 9x^2 + 12x + 1$

Step 3

Bepaal $\frac{dy}{dx}$ indien: $yx - y = 2x^2 - 2x$ ; $x \neq 1$

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Answer

We start with the equation:

$yx - y = 2x^2 - 2x$

Factor the left side: $y(x-1) = 2x^2 - 2x$
Divide by $(x-1)$ (noting $x \neq 1$ ): $y = \frac{2x^2 - 2x}{x-1}$
Differentiate using the quotient rule: $\frac{dy}{dx} = \frac{(4x - 2)(x-1) - (2x^2 - 2x)(1)}{(x-1)^2}$
Simplify:
- First, calculate the numerator: $= (4x - 2)(x - 1) - (2x^2 - 2x)$ $= 4x^2 - 4x - 2x + 2 - 2x^2 + 2x$ $= 2x^2 - 4x + 2$
The final derivative: $\frac{dy}{dx} = \frac{2x^2 - 4x + 2}{(x-1)^2}$

8.1 Bepaal $f'(x)$ vanuit eerste beginsels indien dit gegee word dat $f(x) = x^2 - 5$ - NSC Mathematics - Question 8 - 2018 - Paper 1

Worked Solution & Example Answer:8.1 Bepaal $f'(x)$ vanuit eerste beginsels indien dit gegee word dat $f(x) = x^2 - 5$ - NSC Mathematics - Question 8 - 2018 - Paper 1

Bepaal $f'(x)$ vanuit eerste beginsels indien dit gegee word dat $f(x) = x^2 - 5$

Bepaal $\frac{dy}{dx}$ indien: $y = 3x^3 + 6x^2 + x - 4$

Bepaal $\frac{dy}{dx}$ indien: $yx - y = 2x^2 - 2x$ ; $x \neq 1$

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