Bepaal $f' (x)$ vanuit eerste beginsels as $f (x) = x^2 - 5$ - NSC Mathematics - Question 7 - 2017 - Paper 1
Question 7
Bepaal $f' (x)$ vanuit eerste beginsels as $f (x) = x^2 - 5$.
Bepaal die afgeleide van: $g(x) = 5x^2 - \frac{2x}{x^3}$.
Gegee: $h(x) = a^x, x > 0$.
Bepaal die waar... show full transcript
Worked Solution & Example Answer:Bepaal $f' (x)$ vanuit eerste beginsels as $f (x) = x^2 - 5$ - NSC Mathematics - Question 7 - 2017 - Paper 1
Step 1
Bepaal $f' (x)$ vanuit eerste beginsels
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Answer
To find the derivative f′(x) using first principles, we start with the definition of the derivative:
f′(x)=limh→0hf(x+h)−f(x)
Substituting f(x)=x2−5, we have:
Calculate f(x+h):
f(x+h)=(x+h)2−5=x2+2xh+h2−5
Now compute:
f(x+h)−f(x)=(x2+2xh+h2−5)−(x2−5)=2xh+h2
Substitute this back into the derivative definition:
f′(x)=limh→0h2xh+h2
Simplifying gives:
f′(x)=limh→0(2x+h)
As h approaches 0, we find:
f′(x)=2x
Step 2
Bepaal die afgeleide van: $g(x) = 5x^2 - \frac{2x}{x^3}$
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Answer
To find the derivative g′(x), we first simplify the function:
Rewrite the function:
g(x)=5x2−2x⋅x−3=5x2−2x−2
Differentiate each term:
g′(x)=10x−(−2)x−3=10x+x32
Thus, the derivative is:
g′(x)=10x+x32
Step 3
Bepaal die waarde van $a$ as gegee word dat $h' (8) = h' (4)$
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Answer
We start with the function:
h(x)=ax
The derivative of this function is:
h′(x)=axln(a)
Setting up the condition:
h′(8)=h′(4)
Substituting the expressions:
a8ln(a)=a4ln(a)
Assuming extln(a)eq0, we can simplify by dividing both sides:
a8=a4
This implies:
ightarrow a^4 = 1$$
Therefore:
$$a = 1 ext{ or } a = -1 ext{ (discarded since } a > 0).$$
Thus, the final value of $a$ is:
$$a = 1.$$
