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Bepaal $f^{ ext{'} } (x)$ vanuit eerste beginsels indien dit gegee word dat $f(x)=-x^{2}$ - NSC Mathematics - Question 8 - 2022 - Paper 1

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Bepaal $f^{ ext{'} } (x)$ vanuit eerste beginsels indien dit gegee word dat $f(x)=-x^{2}$. Bepaal: 8.2 $f^{ ext{'} } (x)$, indien dit gegee word dat $f(x)=4x^{3}... show full transcript

Worked Solution & Example Answer:Bepaal $f^{ ext{'} } (x)$ vanuit eerste beginsels indien dit gegee word dat $f(x)=-x^{2}$ - NSC Mathematics - Question 8 - 2022 - Paper 1

Step 1

Bepaal $f^{ ext{'} } (x)$ vanuit eerste beginsels indien dit gegee word dat $f(x)=-x^{2}$.

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Answer

To find the derivative fext′(x)f^{ ext{'} }(x) from first principles, we use the definition of the derivative:

f^{ ext{'} }(x) = rac{f(x+h) - f(x)}{h} Substituting the given function:

f(x)=−x2f(x) = -x^{2} Then,

f(x+h)=−(x+h)2=−x2−2xh−h2f(x+h) = -(x+h)^{2} = -x^{2} - 2xh - h^{2}

Now substituting into the derivative formula:

f^{ ext{'} }(x) = rac{-x^{2} - 2xh - h^{2} + x^{2}}{h} = rac{-2xh - h^{2}}{h} = -2x - h

Taking the limit as hh approaches 0:

fext′(x)=−2xf^{ ext{'} }(x) = -2x

Step 2

Bepaal: 8.2 $f^{ ext{'} } (x)$, indien dit gegee word dat $f(x)=4x^{3}-5x^{2}$.

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Answer

To find the derivative of the function f(x)=4x3−5x2f(x) = 4x^{3} - 5x^{2}, we apply the power rule:

fext′(x)=12x2−10xf^{ ext{'} }(x) = 12x^{2} - 10x

Step 3

8.2.2 $D_{x} \begin{bmatrix} -\frac{6\sqrt{x}+2}{x} \end{bmatrix}$

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Answer

To find the derivative of the function inside the brackets, we use the quotient rule:

Dx(−6x+2x)=−(6⋅12x−1/2)⋅x−(6x+2)(1)x2D_{x} \left( -\frac{6 \sqrt{x} + 2}{x} \right) = -\frac{(6 \cdot \frac{1}{2} x^{-1/2}) \cdot x - (6 \sqrt{x}+2)(1)}{x^{2}}

After simplification, we find:

=−3x1/2⋅x−(6x+2)x2= -\frac{3x^{1/2} \cdot x - (6 \sqrt{x} + 2)}{x^{2}}

Further simplifying gives:

=−3x3/2−6x−2x2= -\frac{3x^{3/2} - 6\sqrt{x} - 2}{x^{2}}

Combining terms leads us to the final result.

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