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Sagot :
Sure, let's go through the solutions step-by-step.
### Part 2.1: Determining [tex]\( f'(x) \)[/tex] from first principles
Given the function:
[tex]\[ f(x) = -2x^2 + 1 \][/tex]
To find the derivative [tex]\( f'(x) \)[/tex] from first principles, we use the definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \][/tex]
First, calculate [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x + h) = -2(x + h)^2 + 1 \][/tex]
[tex]\[ = -2(x^2 + 2xh + h^2) + 1 \][/tex]
[tex]\[ = -2x^2 - 4xh - 2h^2 + 1 \][/tex]
Now, substitute [tex]\( f(x + h) \)[/tex] and [tex]\( f(x) \)[/tex] into the difference quotient:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{(-2x^2 - 4xh - 2h^2 + 1) - (-2x^2 + 1)}{h} \][/tex]
[tex]\[ = \frac{-4xh - 2h^2}{h} \][/tex]
[tex]\[ = -4x - 2h \][/tex]
Taking the limit as [tex]\( h \)[/tex] approaches 0:
[tex]\[ f'(x) = \lim_{h \to 0} (-4x - 2h) = -4x \][/tex]
Thus, the derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = -4x \][/tex]
### Part 2.2: Additional Derivatives
#### Part 2.2.1: Determining [tex]\( f'(x) \)[/tex]
Given the function:
[tex]\[ f(x) = \frac{1}{2} x^2 - \frac{5}{x} \][/tex]
To find the derivative [tex]\( f'(x) \)[/tex], we differentiate each term separately:
[tex]\[ \frac{d}{dx} \left( \frac{1}{2} x^2 \right) = \frac{1}{2} \cdot 2x = x \][/tex]
For the second term, using the power rule where [tex]\( \frac{5}{x} \)[/tex] is rewritten as [tex]\( 5x^{-1} \)[/tex]:
[tex]\[ \frac{d}{dx} \left( - \frac{5}{x} \right) = \frac{d}{dx} (-5x^{-1}) = -5 \cdot (-1)x^{-2} = \frac{5}{x^2} \][/tex]
Combining these results, we get:
[tex]\[ f'(x) = x + \frac{5}{x^2} \][/tex]
### Part 2.2.2: Determining the Derivative of
[tex]\[ g(x) = \frac{-2x^2 + x^{1/4}}{x^2} \][/tex]
First, simplify the function:
[tex]\[ g(x) = \frac{-2x^2}{x^2} + \frac{x^{1/4}}{x^2} \][/tex]
[tex]\[ = -2 + x^{1/4 - 2} \][/tex]
[tex]\[ = -2 + x^{-7/4} \][/tex]
To find the derivative [tex]\( g'(x) \)[/tex], differentiate each term separately:
[tex]\[ \frac{d}{dx} (-2) = 0 \][/tex]
[tex]\[ \frac{d}{dx} (x^{-7/4}) = \left(-\frac{7}{4}\right)x^{-7/4 - 1} = -\frac{7}{4}x^{-11/4} \][/tex]
Thus, the derivative of [tex]\( g(x) \)[/tex] is:
[tex]\[ g'(x) = -\frac{7}{4} x^{-11/4} \][/tex]
To express it back in a more conventional form:
[tex]\[ g'(x) = -\frac{7}{4} \cdot \frac{1}{x^{11/4}} \][/tex]
So summarizing:
2.1: [tex]\( f'(x) = -4x \)[/tex]
2.2.1: [tex]\( f'(x) = x + \frac{5}{x^2} \)[/tex]
2.2.2: [tex]\( D_x\left[ \frac{-2x^2 + \sqrt[4]{x}}{x^2} \right] = -\frac{7}{4} x^{-11/4} \)[/tex]
### Part 2.1: Determining [tex]\( f'(x) \)[/tex] from first principles
Given the function:
[tex]\[ f(x) = -2x^2 + 1 \][/tex]
To find the derivative [tex]\( f'(x) \)[/tex] from first principles, we use the definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \][/tex]
First, calculate [tex]\( f(x + h) \)[/tex]:
[tex]\[ f(x + h) = -2(x + h)^2 + 1 \][/tex]
[tex]\[ = -2(x^2 + 2xh + h^2) + 1 \][/tex]
[tex]\[ = -2x^2 - 4xh - 2h^2 + 1 \][/tex]
Now, substitute [tex]\( f(x + h) \)[/tex] and [tex]\( f(x) \)[/tex] into the difference quotient:
[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{(-2x^2 - 4xh - 2h^2 + 1) - (-2x^2 + 1)}{h} \][/tex]
[tex]\[ = \frac{-4xh - 2h^2}{h} \][/tex]
[tex]\[ = -4x - 2h \][/tex]
Taking the limit as [tex]\( h \)[/tex] approaches 0:
[tex]\[ f'(x) = \lim_{h \to 0} (-4x - 2h) = -4x \][/tex]
Thus, the derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = -4x \][/tex]
### Part 2.2: Additional Derivatives
#### Part 2.2.1: Determining [tex]\( f'(x) \)[/tex]
Given the function:
[tex]\[ f(x) = \frac{1}{2} x^2 - \frac{5}{x} \][/tex]
To find the derivative [tex]\( f'(x) \)[/tex], we differentiate each term separately:
[tex]\[ \frac{d}{dx} \left( \frac{1}{2} x^2 \right) = \frac{1}{2} \cdot 2x = x \][/tex]
For the second term, using the power rule where [tex]\( \frac{5}{x} \)[/tex] is rewritten as [tex]\( 5x^{-1} \)[/tex]:
[tex]\[ \frac{d}{dx} \left( - \frac{5}{x} \right) = \frac{d}{dx} (-5x^{-1}) = -5 \cdot (-1)x^{-2} = \frac{5}{x^2} \][/tex]
Combining these results, we get:
[tex]\[ f'(x) = x + \frac{5}{x^2} \][/tex]
### Part 2.2.2: Determining the Derivative of
[tex]\[ g(x) = \frac{-2x^2 + x^{1/4}}{x^2} \][/tex]
First, simplify the function:
[tex]\[ g(x) = \frac{-2x^2}{x^2} + \frac{x^{1/4}}{x^2} \][/tex]
[tex]\[ = -2 + x^{1/4 - 2} \][/tex]
[tex]\[ = -2 + x^{-7/4} \][/tex]
To find the derivative [tex]\( g'(x) \)[/tex], differentiate each term separately:
[tex]\[ \frac{d}{dx} (-2) = 0 \][/tex]
[tex]\[ \frac{d}{dx} (x^{-7/4}) = \left(-\frac{7}{4}\right)x^{-7/4 - 1} = -\frac{7}{4}x^{-11/4} \][/tex]
Thus, the derivative of [tex]\( g(x) \)[/tex] is:
[tex]\[ g'(x) = -\frac{7}{4} x^{-11/4} \][/tex]
To express it back in a more conventional form:
[tex]\[ g'(x) = -\frac{7}{4} \cdot \frac{1}{x^{11/4}} \][/tex]
So summarizing:
2.1: [tex]\( f'(x) = -4x \)[/tex]
2.2.1: [tex]\( f'(x) = x + \frac{5}{x^2} \)[/tex]
2.2.2: [tex]\( D_x\left[ \frac{-2x^2 + \sqrt[4]{x}}{x^2} \right] = -\frac{7}{4} x^{-11/4} \)[/tex]
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