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8.1 Geege: $f(x) = -2x^2 + p$ Bepaal $f'(x)$ vanuit eerste beginsels - NSC Mathematics - Question 8 - 2017 - Paper 1

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8.1-Geege:-$f(x)-=--2x^2-+-p$--Bepaal-$f'(x)$-vanuit-eerste-beginsels-NSC Mathematics-Question 8-2017-Paper 1.png

8.1 Geege: $f(x) = -2x^2 + p$ Bepaal $f'(x)$ vanuit eerste beginsels. 8.2 Bepaal: $D \left[ 4 \sqrt{x} + \frac{1}{3x} + 2 \right]$

Worked Solution & Example Answer:8.1 Geege: $f(x) = -2x^2 + p$ Bepaal $f'(x)$ vanuit eerste beginsels - NSC Mathematics - Question 8 - 2017 - Paper 1

Step 1

Bepaal $f'(x)$ vanuit eerste beginsels

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Answer

Om f′(x)f'(x) te berekenen vanuit eerste beginsels, gebruiken we de limietdefinitie van de afgeleide:

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

Substitueer f(x)f(x) in de formule:

f′(x)=lim⁡h→0−2(x+h)2+p−(−2x2+p)hf'(x) = \lim_{h \to 0} \frac{-2(x+h)^2 + p - (-2x^2 + p)}{h}

Vereenvoudig dit:

=lim⁡h→0−2(x2+2xh+h2)+p+2x2−ph= \lim_{h \to 0} \frac{-2(x^2 + 2xh + h^2) + p + 2x^2 - p}{h}

=lim⁡h→0−4xh−2h2h= \lim_{h \to 0} \frac{-4xh - 2h^2}{h}

=lim⁡h→0−4x−2h= \lim_{h \to 0} -4x - 2h

Wanneer we hh laten toenemen naar 0, krijgen we:

f′(x)=−4xf'(x) = -4x

Step 2

Bepaal: $D \left[ 4 \sqrt{x} + \frac{1}{3x} + 2 \right]$

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Answer

Om de afgeleide DD te bepalen, passen we de afgeleide regels toe:

D=4⋅12x−13x2D = 4 \cdot \frac{1}{2\sqrt{x}} - \frac{1}{3x^2}

Dit geeft ons:

D=2x−13x2D = \frac{2}{\sqrt{x}} - \frac{1}{3x^2}

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