Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = -3x$. Determine: $6.2.1 \ D_x\left[ p^3x^2 - 7x + 10 \right]$ $6.2.2 \ dy \ dx$ if $y = \frac{x - 3x^2}{2}$. $6.2.3 \ f'(x)$ if $f(x) = \sqrt{x} + 5x^4$. The equation of the tangent defined by $f(x)=mx+c$ to the curve $g(x)=x^2 + 3x - 2$, is perpendicular to the line $p(x) = \frac{1}{9}x - 4$. $6.3.1$ Write down the value of $m$. $6.3.2$ Hence, determine the coordinates of the point of contact of the tangent to the curve $g$. $6.3.3$ Determine the average gradient of $g$ between the points where $x = 3$ and $x = 2$.

NSC Technical Mathematics Differential Calculus and Integration: Determine $f'(x)$ using FIRST P

Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = -3x$ - NSC Technical Mathematics - Question 6 - 2021 - Paper 1

Question 6

Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = -3x$. Determine: $6.2.1 \ D_x\left[ p^3x^2 - 7x + 10 \right]$ $6.2.2 \ dy \ dx$ if $y = \frac{x - 3x^2}{2}$. ... show full transcript

Worked Solution & Example Answer:Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = -3x$ - NSC Technical Mathematics - Question 6 - 2021 - Paper 1

Step 1

Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = -3x$

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Answer

To determine the derivative using first principles, we apply the definition:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Substituting for $f(x)$ :

$= \lim_{h \to 0} \frac{-3(x+h) - (-3x)}{h} = \lim_{h \to 0} \frac{-3x - 3h + 3x}{h} = \lim_{h \ to 0} \frac{-3h}{h} = \lim_{h \ to 0} -3 = -3$

Thus, $f'(x) = -3$ .

Step 2

Determine: $6.2.1 \ D_x\left[ p^3x^2 - 7x + 10 \right]$

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Answer

To find the first derivative, apply the power rule:

$D_x [p^3x^2 - 7x + 10] = 2p^3x - 7$

Step 3

Determine: $6.2.2 \ dy \ dx$ if $y = \frac{x - 3x^2}{2}$

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Answer

First, differentiate with respect to $x$ :

$y = \frac{x - 3x^2}{2}$

$dy \ dx = \frac{1 - 6x}{2}$

Step 4

Determine: $6.2.3 \ f'(x)$ if $f(x) = \sqrt{x} + 5x^4$

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Answer

To differentiate $f(x)$ , use the power rule and the derivative of square root:

$f'(x) = \frac{1}{2\sqrt{x}} + 20x^3$

Step 5

Determine the equation of the tangent defined by $f(x)=mx+c$ to the curve $g(x)=x^2 + 3x - 2$, is perpendicular to the line $p(x) = \frac{1}{9}x - 4$

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Answer

The slope of the line $p(x)$ is $rac{1}{9}$ . Since the tangent is perpendicular, the slope $m$ can be found by:

$m = -9$

Step 6

Determine: $6.3.2$ Hence, determine the coordinates of the point of contact of the tangent to the curve $g$

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Answer

Set the derivative of $g(x)$ equal to the slope at the tangent point:

$g'(x) = 2x + 3$

Setting $m = -9$ :

$2x + 3 = -9\Rightarrow x = -6$

Substituting $x$ into $g(x)$ :

$g(-6) = (-6)^2 + 3(-6) - 2 = 16$

Therefore, the point of contact is $(-6, 16)$ .

Step 7

Determine: $6.3.3$ Determine the average gradient of $g$ between the points where $x = 3$ and $x = 2$

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Answer

Calculate the average gradient using the formula:

$\text{Average gradient} = \frac{g(x_2) - g(x_1)}{x_2 - x_1}$

Substituting $x_1 = 2$ and $x_2 = 3$ :

$g(3) = 16 \quad g(2) = 4$

Thus:

$\text{Average gradient} = \frac{16 - 4}{3 - 2} = 12$

Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = -3x$ - NSC Technical Mathematics - Question 6 - 2021 - Paper 1

Worked Solution & Example Answer:Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = -3x$ - NSC Technical Mathematics - Question 6 - 2021 - Paper 1

Determine $f'(x)$ using FIRST PRINCIPLES if $f(x) = -3x$

Determine: $6.2.1 \ D_x\left[ p^3x^2 - 7x + 10 \right]$

Determine: $6.2.2 \ dy \ dx$ if $y = \frac{x - 3x^2}{2}$

Determine: $6.2.3 \ f'(x)$ if $f(x) = \sqrt{x} + 5x^4$

Determine the equation of the tangent defined by $f(x)=mx+c$ to the curve $g(x)=x^2 + 3x - 2$, is perpendicular to the line $p(x) = \frac{1}{9}x - 4$

Determine: $6.3.2$ Hence, determine the coordinates of the point of contact of the tangent to the curve $g$

Determine: $6.3.3$ Determine the average gradient of $g$ between the points where $x = 3$ and $x = 2$

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