Given: $f(x) = x - 5 $ Determine $f'(x)$ using FIRST PRINCIPLES. --- 6.2 Determine: 6.2.1 $D_{x}[-3x^{2} - 7x]$ 6.2.2 $f'(x)$ if $f(x) = \frac{3}{2}x + \sqrt{x^{2}}$ 6.2.3 $rac{dy}{dt}$ if $y^{3} = 64$ --- Given: $h(x) = -2x^{2} + x - 5$ 6.3.1 Calculate $h(1)$. 6.3.2 Hence, determine the average gradient of $h$ between the points $(1; h(1))$ and $(-3; -26)$. --- 6.4 Determine the equation of the tangent to the curve defined by $f(x) = x^{3} + 2$ at $x = 4$.

NSC Technical Mathematics Differential Calculus and Integration: Given: $f(x) = x

Given: $f(x) = x - 5 $ Determine $f'(x)$ using FIRST PRINCIPLES - NSC Technical Mathematics - Question 6 - 2023 - Paper 1

Question 6

Given: $f(x) = x - 5 $ Determine $f'(x)$ using FIRST PRINCIPLES. --- 6.2 Determine: 6.2.1 $D_{x}[-3x^{2} - 7x]$ 6.2.2 $f'(x)$ if $f(x) = \frac{3}{2}x + \sqrt{... show full transcript

Worked Solution & Example Answer:Given: $f(x) = x - 5 $ Determine $f'(x)$ using FIRST PRINCIPLES - NSC Technical Mathematics - Question 6 - 2023 - Paper 1

Step 1

Determine $f'(x)$ using FIRST PRINCIPLES.

96%

114 rated

Answer

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

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

In our case, we substitute:

[ f(x+h) = (x+h) - 5 ] [ f(x) = x - 5 ]

Now, applying the definition, we get:

[ f'(x) = \lim_{h \to 0} \frac{(x + h - 5) - (x - 5)}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1. ]

Thus, the derivative is:

$f'(x) = 1$ .

Step 2

Determine $D_{x}[-3x^{2} - 7x]$

99%

104 rated

Answer

To find the derivative of the function, we differentiate with respect to $x$ :

[ D_{x}[-3x^{2} - 7x] = -6x - 7.]

Thus, the answer is:

$D_{x}[-3x^{2} - 7x] = -6x - 7$ .

Step 3

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

96%

101 rated

Answer

First, rewrite the function:

[ f(x) = \frac{3}{2}x + x. ]

Now, differentiate:

[ f'(x) = \frac{3}{2} + 1 = \frac{5}{2}. ]

Thus, the answer is:

$f'(x) = \frac{5}{2}$ .

Step 4

$\frac{dy}{dt}$ if $y^{3} = 64$

98%

120 rated

Answer

First, solve for $y$ :

[ y = 4. ]

Differentiating implicitly, we have:

[ 3y^{2} \frac{dy}{dt} = 0 \implies \frac{dy}{dt} = \frac{0}{3y^{2}}. ]

Since $y = 4$ , then:

[ \frac{dy}{dt} = 0. ]

Thus, the answer is:

$\frac{dy}{dt} = 0$ .

Step 5

Calculate $h(1)$.

97%

117 rated

Answer

To calculate $h(1)$ , substitute $x = 1$ into the equation:

[ h(1) = -2(1)^{2} + (1) - 5 = -2 + 1 - 5 = -6. ]

Thus, the answer is:

$h(1) = -6$ .

Step 6

Hence, determine the average gradient of $h$ between the points $(1; h(1))$ and $(-3; -26)$.

97%

121 rated

Answer

The average gradient is calculated as follows:

[ \text{Gradient} = \frac{h(x_{2}) - h(x_{1})}{x_{2} - x_{1}} = \frac{-26 - (-6)}{-3 - 1} = \frac{-26 + 6}{-4} = \frac{-20}{-4} = 5. ]

Thus, the answer is:

$5$ .

Step 7

Determine the equation of the tangent to the curve defined by $f(x) = x^{3} + 2$ at $x = 4$.

96%

114 rated

Answer

First, find $f'(x)$ :

[ f'(x) = 3x^{2} \implies f'(4) = 3(4)^{2} = 48. ]

Now, using the point-slope form of the line:

[ y - f(4) = m(x - 4) ]

To find $f(4)$ :

[ f(4) = 4^{3} + 2 = 66. ]

This gives us the equation:

[ y - 66 = 48(x - 4) \Rightarrow y = 48x - 192 + 66 \Rightarrow y = 48x - 126. ]

Thus, the equation of the tangent is:

$y = 48x - 126$ .

Given: $f(x) = x - 5 $ Determine $f'(x)$ using FIRST PRINCIPLES - NSC Technical Mathematics - Question 6 - 2023 - Paper 1

Worked Solution & Example Answer:Given: $f(x) = x - 5 $ Determine $f'(x)$ using FIRST PRINCIPLES - NSC Technical Mathematics - Question 6 - 2023 - Paper 1

Determine $f'(x)$ using FIRST PRINCIPLES.

Determine $D_{x}[-3x^{2} - 7x]$

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

$\frac{dy}{dt}$ if $y^{3} = 64$

Calculate $h(1)$.

Hence, determine the average gradient of $h$ between the points $(1; h(1))$ and $(-3; -26)$.

Determine the equation of the tangent to the curve defined by $f(x) = x^{3} + 2$ at $x = 4$.

Join the NSC students using SimpleStudy...