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7.1 Bepaal $f'(x)$ vanuit eerste beginsels indien $f(x)=2x^2-1$ - NSC Mathematics - Question 7 - 2020 - Paper 1

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7.1 Bepaal $f'(x)$ vanuit eerste beginsels indien $f(x)=2x^2-1$. 7.2 Bepaal: 7.2.1 $\frac{d}{dx}(\sqrt{x^2+x^3})$ 7.2.2 $f'(x)$ as $f(x)=\frac{4x^2-9}{4x+6}; \, x... show full transcript

Worked Solution & Example Answer:7.1 Bepaal $f'(x)$ vanuit eerste beginsels indien $f(x)=2x^2-1$ - NSC Mathematics - Question 7 - 2020 - Paper 1

Step 1

Bepaal $f'(x)$ vanuit eerste beginsels indien $f(x)=2x^2-1$

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Answer

To find the derivative using first principles, we will use the limit definition of the derivative:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  1. Calculate f(x+h)f(x+h):
    f(x+h)=2(x+h)2−1=2(x2+2xh+h2)−1=2x2+4xh+2h2−1f(x+h) = 2(x+h)^2 - 1 = 2(x^2 + 2xh + h^2) - 1 = 2x^2 + 4xh + 2h^2 - 1

  2. Find f(x+h)−f(x)f(x+h) - f(x):
    f(x+h)−f(x)=(2x2+4xh+2h2−1)−(2x2−1)=4xh+2h2f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 1) - (2x^2 - 1) = 4xh + 2h^2

  3. Substitute into the limit:
    f′(x)=lim⁡h→04xh+2h2h=lim⁡h→0(4x+2h)=4xf'(x) = \lim_{h \to 0} \frac{4xh + 2h^2}{h} = \lim_{h \to 0} (4x + 2h) = 4x

Thus, the derivative is:
f′(x)=4xf'(x) = 4x

Step 2

Bepaal: $\frac{d}{dx}(\sqrt{x^2+x^3})$

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Answer

To differentiate x2+x3\sqrt{x^2 + x^3}, we can use the chain rule. Let u=x2+x3u = x^2 + x^3, then:

  1. Differentiate using chain rule:
    ddx(u)=12u⋅dudx\frac{d}{dx}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}

  2. Calculate dudx\frac{du}{dx}:
    dudx=2x+3x2\frac{du}{dx} = 2x + 3x^2

  3. Therefore,
    ddx(x2+x3)=12x2+x3⋅(2x+3x2)\frac{d}{dx}(\sqrt{x^2+x^3}) = \frac{1}{2\sqrt{x^2+x^3}} \cdot (2x + 3x^2)
    The final answer is
    2x+3x22x2+x3\frac{2x + 3x^2}{2\sqrt{x^2 + x^3}}

Step 3

Bepaal $f'(x)$ as $f(x)=\frac{4x^2-9}{4x+6}; \, x=\frac{3}{2}$

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Answer

To find the derivative of f(x)=4x2−94x+6f(x) = \frac{4x^2 - 9}{4x + 6}, we can use the quotient rule:

  1. Differentiate using the quotient rule:
    f′(x)=(g(x)⋅f′(x)−f(x)⋅g′(x))(g(x))2f'(x) = \frac{(g(x) \cdot f'(x) - f(x) \cdot g'(x))}{(g(x))^2}
    where f(x)=4x2−9f(x) = 4x^2 - 9 and g(x)=4x+6g(x) = 4x + 6.

  2. Compute f′(x)f'(x) and g′(x)g'(x):

    • f′(x)=8xf'(x) = 8x
    • g′(x)=4g'(x) = 4

  3. Applying the quotient rule gives: f′(x)=(4x+6)(8x)−(4x2−9)(4)(4x+6)2f'(x) = \frac{(4x + 6)(8x) - (4x^2 - 9)(4)}{(4x + 6)^2}

  4. To find f′(32)f'(\frac{3}{2}), substitute x=32x = \frac{3}{2}: f′(32)=(4⋅32+6)(8⋅32)−(4(32)2−9)(4)(4⋅32+6)2f'\left(\frac{3}{2}\right) = \frac{(4 \cdot \frac{3}{2} + 6)(8 \cdot \frac{3}{2}) - (4\left(\frac{3}{2}\right)^2 - 9)(4)}{(4 \cdot \frac{3}{2} + 6)^2}

  5. Calculate to find the answer.

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