Photo AI

Bepaal $f'(x)$ vanuit eerste beginsels as gegee word dat $f(x) = 2x^2 - 3x$ - NSC Mathematics - Question 9 - 2021 - Paper 1

Question icon

Question 9

Bepaal--$f'(x)$--vanuit-eerste-beginsels-as-gegee-word-dat--$f(x)-=-2x^2---3x$-NSC Mathematics-Question 9-2021-Paper 1.png

Bepaal $f'(x)$ vanuit eerste beginsels as gegee word dat $f(x) = 2x^2 - 3x$. 9.2 Bepaal: 9.2.1 $\frac{dy}{dx}$ as $y = 4x^4 - 6x^3 + 3x$ 9.2.2 $D_x\le... show full transcript

Worked Solution & Example Answer:Bepaal $f'(x)$ vanuit eerste beginsels as gegee word dat $f(x) = 2x^2 - 3x$ - NSC Mathematics - Question 9 - 2021 - Paper 1

Step 1

Bepaal $f'(x)$ vanuit eerste beginsels as gegee word dat $f(x) = 2x^2 - 3x$

96%

114 rated

Answer

To find the derivative from first principles, we use the 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. Substitute the function into the formula:

    • First, calculate f(x+h)f(x + h): f(x+h)=2(x+h)2−3(x+h)f(x + h) = 2(x + h)^2 - 3(x + h)
    • Expanding gives: f(x+h)=2(x2+2xh+h2)−3x−3hf(x + h) = 2(x^2 + 2xh + h^2) - 3x - 3h =2x2+4xh+2h2−3x−3h= 2x^2 + 4xh + 2h^2 - 3x - 3h

  2. Now calculate f(x+h)−f(x)f(x + h) - f(x): f(x+h)−f(x)=(2x2+4xh+2h2−3x−3h)−(2x2−3x)f(x + h) - f(x) = (2x^2 + 4xh + 2h^2 - 3x - 3h) - (2x^2 - 3x) =4xh+2h2−3h= 4xh + 2h^2 - 3h

  3. Now place into the limit formula: f′(x)=lim⁡h→04xh+2h2−3hhf'(x) = \lim_{h \to 0} \frac{4xh + 2h^2 - 3h}{h} =lim⁡h→0(4x+2h−3)= \lim_{h \to 0} (4x + 2h - 3)

  4. Evaluate the limit as hh approaches 0: f′(x)=4x−3f'(x) = 4x - 3

Step 2

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

99%

104 rated

Answer

To find \frac{dy}{dx}$$, we differentiate the function with respect to x$:

dydx=ddx(4x4−6x3+3x)\frac{dy}{dx} = \frac{d}{dx}(4x^4 - 6x^3 + 3x)

Using the power rule, we get:

  1. Differentiate 4x44x^4: [4 \cdot 4x^{4-1} = 16x^3]
  2. Differentiate −6x3-6x^3: [-6 \cdot 3x^{3-1} = -18x^2]
  3. Differentiate 3x3x: [3 \cdot 1 = 3]

Putting it all together, we have:

dydx=16x3−18x2+3\frac{dy}{dx} = 16x^3 - 18x^2 + 3

Step 3

Bepaal: $D_x\left[ \sqrt{\frac{x}{2}} + \left( \frac{1}{3x} \right)^2 \right]$

96%

101 rated

Answer

To find the derivative DxD_x, we differentiate the expression step by step:

First, denote the function as: z=x2+(13x)2z = \sqrt{\frac{x}{2}} + \left( \frac{1}{3x} \right)^2

  1. Differentiate x2\sqrt{\frac{x}{2}}:

    • Using the chain rule: ddx(x2)=12x2⋅12=12x2⋅12=14x2\frac{d}{dx}\left(\sqrt{\frac{x}{2}}\right) = \frac{1}{2\sqrt{\frac{x}{2}}} \cdot \frac{1}{2} = \frac{1}{2 \sqrt{\frac{x}{2}}}\cdot \frac{1}{2} = \frac{1}{4\sqrt{\frac{x}{2}}}

  2. Differentiate (13x)2\left( \frac{1}{3x} \right)^2:

    • First, identify it as 19x2\frac{1}{9x^2}: ddx(19x2)=−29x3\frac{d}{dx}\left(\frac{1}{9x^2}\right) = -\frac{2}{9x^3}

Combining these results gives:

Dxz=14x2−29x3D_x z = \frac{1}{4\sqrt{\frac{x}{2}}} - \frac{2}{9x^3}

Join the NSC students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;