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7.1 Determine $f^{ ext{'}}(x)$ from first principles if $f(x) = x^2 - 5$ - NSC Mathematics - Question 7 - 2017 - Paper 1

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7.1-Determine-$f^{-ext{'}}(x)$-from-first-principles-if-$f(x)-=-x^2---5$-NSC Mathematics-Question 7-2017-Paper 1.png

7.1 Determine $f^{ ext{'}}(x)$ from first principles if $f(x) = x^2 - 5$. 7.2 Determine the derivative of: $g(x) = 5x^3 - \frac{2x}{x^3}$. 7.3 Given: $h(x) = a x^2... show full transcript

Worked Solution & Example Answer:7.1 Determine $f^{ ext{'}}(x)$ from first principles if $f(x) = x^2 - 5$ - NSC Mathematics - Question 7 - 2017 - Paper 1

Step 1

Determine $f^{\text{'}}(x)$ from first principles if $f(x) = x^2 - 5$

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Answer

To find the derivative from first principles, we use the formula:

f(x)=limh0f(x+h)f(x)hf^{\text{'}}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

First, we compute f(x+h)f(x + h):

f(x+h)=(x+h)25=(x2+2xh+h25)f(x + h) = (x + h)^2 - 5 = (x^2 + 2xh + h^2 - 5)

Now, substituting into the derivative formula:

f(x)=limh0(x2+2xh+h25)(x25)hf^{\text{'}}(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 - 5) - (x^2 - 5)}{h}

Simplifying, we obtain:

f(x)=limh02xh+h2h=limh0(2x+h)=2xf^{\text{'}}(x) = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x

Step 2

Determine the derivative of: $g(x) = 5x^3 - \frac{2x}{x^3}$

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Answer

We first simplify the function:

g(x)=5x32x13=5x32x2g(x) = 5x^3 - 2x^{1-3} = 5x^3 - 2x^{-2}

Now, we differentiate using the power rule:

g(x)=15x2+4x3=15x2+4x3g^{\text{'}}(x) = 15x^2 + 4x^{-3} = 15x^2 + \frac{4}{x^3}

Step 3

Given: $h(x) = ax^2, x > 0$. Determine the value of $a$ if it is given that $h^{\text{'}}(8) = h^{\text{'}}(4)$

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Answer

First, we compute the derivative:

h(x)=2axh^{\text{'}}(x) = 2ax

Now, substituting values into the equation:

h(8)=2a(8)=16ah^{\text{'}}(8) = 2a(8) = 16a h(4)=2a(4)=8ah^{\text{'}}(4) = 2a(4) = 8a

Setting them equal:

16a=8a16a = 8a

Solving:

16a8a=08a=0a=1816a - 8a = 0 \Rightarrow 8a = 0 \Rightarrow a = \frac{1}{8}

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