8.1 Determine f '(x) from first principles if f (x) = \frac{1}{x} - NSC Mathematics - Question 8 - 2024 - Paper 1

To differentiate the expression (\sqrt{4x^2 + \sqrt{2x^2}}), we will use the chain rule:

1. Differentiate the outer function:

$\frac{1}{2\sqrt{4x^2 + \sqrt{2x^2}}}$

2. Differentiate the inner function, which involves both terms:

   • The derivative of (4x^2) is (8x).
   • The derivative of (\sqrt{2x^2}) is (\frac{2x}{\sqrt{2x^2}} = \frac{2x}{\sqrt{2} \cdot x} = \frac{2}{\sqrt{2}}). So:

3. Combine the derivatives:

$\frac{1}{2\sqrt{4x^2 + \sqrt{2x^2}}}(8x + \frac{2}{\sqrt{2}})$

Thus:

$\frac{d}{dx}(\sqrt{4x^2 + \sqrt{2x^2}}) = \frac{2x}{\sqrt{4x^2 + \sqrt{2x^2}}} + \frac{1}{2\sqrt{2}(\sqrt{4x^2 + \sqrt{2x^2}})}.$

To simplify g(x):

$g(x) = \frac{3x^4 - 4x^2 + 6}{x^2}$

This can be separated into terms:

$g(x) = 3x^2 - 4 + \frac{6}{x^2}.$

Thus, the simplified form of g(x) is:

$g(x) = 3x^2 - 4 + 6x^{-2}.$

To find b and c, we need the following:

1. Calculate f '(x) at x = 1:

   • For (f(x) = 3x^3 + bx + c),
   • The derivative is (f'(x) = 9x^2 + b).
   • Evaluating at (x = 1), we have: $f'(1) = 9(1)^2 + b = 9 + b$.

2. Set it equal to the slope of the tangent line:

   • The slope of y = 9x - 9 is 9: $9 + b = 9$ Thus, (b = 0).

3. Calculate f(1):

   • We need the function value at x = 1: $f(1) = 3(1)^3 + 0 + c = 3 + c$
   • Set it equal to y = 9(1) - 9: $3 + c = 0$ Thus, (c = -3).

Finally, the values are:

• (b = 0)
• (c = -3).