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Sagot :
Certainly! Let's go through each differentiation step by step:
### Example 1: Differentiate with respect to [tex]\(x\)[/tex]
#### i) [tex]\(3x^2 + 4x\)[/tex]
To differentiate [tex]\(3x^2 + 4x\)[/tex] with respect to [tex]\(x\)[/tex]:
- Differentiate [tex]\(3x^2\)[/tex]:
[tex]\[ \frac{d}{dx}(3x^2) = 3 \cdot 2x = 6x \][/tex]
- Differentiate [tex]\(4x\)[/tex]:
[tex]\[ \frac{d}{dx}(4x) = 4 \][/tex]
Thus, the derivative of [tex]\(3x^2 + 4x\)[/tex] is:
[tex]\[ 6x + 4 \][/tex]
#### ii) [tex]\(\frac{3}{x^2} + 5x + 6\)[/tex]
To differentiate [tex]\(\frac{3}{x^2} + 5x + 6\)[/tex] with respect to [tex]\(x\)[/tex]:
- Rewrite [tex]\(\frac{3}{x^2}\)[/tex] as [tex]\(3x^{-2}\)[/tex].
- Differentiate [tex]\(3x^{-2}\)[/tex]:
[tex]\[ \frac{d}{dx}(3x^{-2}) = 3 \cdot (-2)x^{-3} = -6x^{-3} = -\frac{6}{x^3} \][/tex]
- Differentiate [tex]\(5x\)[/tex]:
[tex]\[ \frac{d}{dx}(5x) = 5 \][/tex]
- Differentiate [tex]\(6\)[/tex]:
[tex]\[ \frac{d}{dx}(6) = 0 \][/tex]
Thus, the derivative of [tex]\(\frac{3}{x^2} + 5x + 6\)[/tex] is:
[tex]\[ -\frac{6}{x^3} + 5 \][/tex]
Rewriting this in a standard format, we get:
[tex]\[ 5 - \frac{6}{x^3} \][/tex]
#### iii) [tex]\(5x^2 - 4x + 8\)[/tex]
To differentiate [tex]\(5x^2 - 4x + 8\)[/tex] with respect to [tex]\(x\)[/tex]:
- Differentiate [tex]\(5x^2\)[/tex]:
[tex]\[ \frac{d}{dx}(5x^2) = 5 \cdot 2x = 10x \][/tex]
- Differentiate [tex]\(-4x\)[/tex]:
[tex]\[ \frac{d}{dx}(-4x) = -4 \][/tex]
- Differentiate [tex]\(8\)[/tex]:
[tex]\[ \frac{d}{dx}(8) = 0 \][/tex]
Thus, the derivative of [tex]\(5x^2 - 4x + 8\)[/tex] is:
[tex]\[ 10x - 4 \][/tex]
#### iv) [tex]\(4x^3 - 3x^2 + \frac{5}{x^2}\)[/tex]
To differentiate [tex]\(4x^3 - 3x^2 + \frac{5}{x^2}\)[/tex] with respect to [tex]\(x\)[/tex]:
- Differentiate [tex]\(4x^3\)[/tex]:
[tex]\[ \frac{d}{dx}(4x^3) = 4 \cdot 3x^2 = 12x^2 \][/tex]
- Differentiate [tex]\(-3x^2\)[/tex]:
[tex]\[ \frac{d}{dx}(-3x^2) = -3 \cdot 2x = -6x \][/tex]
- Rewrite [tex]\(\frac{5}{x^2}\)[/tex] as [tex]\(5x^{-2}\)[/tex].
- Differentiate [tex]\(5x^{-2}\)[/tex]:
[tex]\[ \frac{d}{dx}(5x^{-2}) = 5 \cdot (-2)x^{-3} = -10x^{-3} = -\frac{10}{x^3} \][/tex]
Thus, the derivative of [tex]\(4x^3 - 3x^2 + \frac{5}{x^2}\)[/tex] is:
[tex]\[ 12x^2 - 6x - \frac{10}{x^3} \][/tex]
#### v) [tex]\(\frac{1}{2}x^2 - \frac{1}{2x^2}\)[/tex]
To differentiate [tex]\(\frac{1}{2}x^2 - \frac{1}{2x^2}\)[/tex] with respect to [tex]\(x\)[/tex]:
- Differentiate [tex]\(\frac{1}{2}x^2\)[/tex]:
[tex]\[ \frac{d}{dx}\left(\frac{1}{2}x^2\right) = \frac{1}{2} \cdot 2x = x \][/tex]
- Rewrite [tex]\(-\frac{1}{2x^2}\)[/tex] as [tex]\(-\frac{1}{2}x^{-2}\)[/tex].
- Differentiate [tex]\(-\frac{1}{2}x^{-2}\)[/tex]:
[tex]\[ \frac{d}{dx}\left(-\frac{1}{2}x^{-2}\right) = -\frac{1}{2} \cdot (-2)x^{-3} = x^{-3} = \frac{1}{x^3} \][/tex]
Thus, the derivative of [tex]\(\frac{1}{2}x^2 - \frac{1}{2x^2}\)[/tex] is:
[tex]\[ x + \frac{1}{x^3} \][/tex]
Putting it all together, we have the final derivatives:
i) [tex]\(6x + 4\)[/tex]
ii) [tex]\(5 - \frac{6}{x^3}\)[/tex]
iii) [tex]\(10x - 4\)[/tex]
iv) [tex]\(12x^2 - 6x - \frac{10}{x^3}\)[/tex]
v) [tex]\(x + \frac{1}{x^3}\)[/tex]
### Example 1: Differentiate with respect to [tex]\(x\)[/tex]
#### i) [tex]\(3x^2 + 4x\)[/tex]
To differentiate [tex]\(3x^2 + 4x\)[/tex] with respect to [tex]\(x\)[/tex]:
- Differentiate [tex]\(3x^2\)[/tex]:
[tex]\[ \frac{d}{dx}(3x^2) = 3 \cdot 2x = 6x \][/tex]
- Differentiate [tex]\(4x\)[/tex]:
[tex]\[ \frac{d}{dx}(4x) = 4 \][/tex]
Thus, the derivative of [tex]\(3x^2 + 4x\)[/tex] is:
[tex]\[ 6x + 4 \][/tex]
#### ii) [tex]\(\frac{3}{x^2} + 5x + 6\)[/tex]
To differentiate [tex]\(\frac{3}{x^2} + 5x + 6\)[/tex] with respect to [tex]\(x\)[/tex]:
- Rewrite [tex]\(\frac{3}{x^2}\)[/tex] as [tex]\(3x^{-2}\)[/tex].
- Differentiate [tex]\(3x^{-2}\)[/tex]:
[tex]\[ \frac{d}{dx}(3x^{-2}) = 3 \cdot (-2)x^{-3} = -6x^{-3} = -\frac{6}{x^3} \][/tex]
- Differentiate [tex]\(5x\)[/tex]:
[tex]\[ \frac{d}{dx}(5x) = 5 \][/tex]
- Differentiate [tex]\(6\)[/tex]:
[tex]\[ \frac{d}{dx}(6) = 0 \][/tex]
Thus, the derivative of [tex]\(\frac{3}{x^2} + 5x + 6\)[/tex] is:
[tex]\[ -\frac{6}{x^3} + 5 \][/tex]
Rewriting this in a standard format, we get:
[tex]\[ 5 - \frac{6}{x^3} \][/tex]
#### iii) [tex]\(5x^2 - 4x + 8\)[/tex]
To differentiate [tex]\(5x^2 - 4x + 8\)[/tex] with respect to [tex]\(x\)[/tex]:
- Differentiate [tex]\(5x^2\)[/tex]:
[tex]\[ \frac{d}{dx}(5x^2) = 5 \cdot 2x = 10x \][/tex]
- Differentiate [tex]\(-4x\)[/tex]:
[tex]\[ \frac{d}{dx}(-4x) = -4 \][/tex]
- Differentiate [tex]\(8\)[/tex]:
[tex]\[ \frac{d}{dx}(8) = 0 \][/tex]
Thus, the derivative of [tex]\(5x^2 - 4x + 8\)[/tex] is:
[tex]\[ 10x - 4 \][/tex]
#### iv) [tex]\(4x^3 - 3x^2 + \frac{5}{x^2}\)[/tex]
To differentiate [tex]\(4x^3 - 3x^2 + \frac{5}{x^2}\)[/tex] with respect to [tex]\(x\)[/tex]:
- Differentiate [tex]\(4x^3\)[/tex]:
[tex]\[ \frac{d}{dx}(4x^3) = 4 \cdot 3x^2 = 12x^2 \][/tex]
- Differentiate [tex]\(-3x^2\)[/tex]:
[tex]\[ \frac{d}{dx}(-3x^2) = -3 \cdot 2x = -6x \][/tex]
- Rewrite [tex]\(\frac{5}{x^2}\)[/tex] as [tex]\(5x^{-2}\)[/tex].
- Differentiate [tex]\(5x^{-2}\)[/tex]:
[tex]\[ \frac{d}{dx}(5x^{-2}) = 5 \cdot (-2)x^{-3} = -10x^{-3} = -\frac{10}{x^3} \][/tex]
Thus, the derivative of [tex]\(4x^3 - 3x^2 + \frac{5}{x^2}\)[/tex] is:
[tex]\[ 12x^2 - 6x - \frac{10}{x^3} \][/tex]
#### v) [tex]\(\frac{1}{2}x^2 - \frac{1}{2x^2}\)[/tex]
To differentiate [tex]\(\frac{1}{2}x^2 - \frac{1}{2x^2}\)[/tex] with respect to [tex]\(x\)[/tex]:
- Differentiate [tex]\(\frac{1}{2}x^2\)[/tex]:
[tex]\[ \frac{d}{dx}\left(\frac{1}{2}x^2\right) = \frac{1}{2} \cdot 2x = x \][/tex]
- Rewrite [tex]\(-\frac{1}{2x^2}\)[/tex] as [tex]\(-\frac{1}{2}x^{-2}\)[/tex].
- Differentiate [tex]\(-\frac{1}{2}x^{-2}\)[/tex]:
[tex]\[ \frac{d}{dx}\left(-\frac{1}{2}x^{-2}\right) = -\frac{1}{2} \cdot (-2)x^{-3} = x^{-3} = \frac{1}{x^3} \][/tex]
Thus, the derivative of [tex]\(\frac{1}{2}x^2 - \frac{1}{2x^2}\)[/tex] is:
[tex]\[ x + \frac{1}{x^3} \][/tex]
Putting it all together, we have the final derivatives:
i) [tex]\(6x + 4\)[/tex]
ii) [tex]\(5 - \frac{6}{x^3}\)[/tex]
iii) [tex]\(10x - 4\)[/tex]
iv) [tex]\(12x^2 - 6x - \frac{10}{x^3}\)[/tex]
v) [tex]\(x + \frac{1}{x^3}\)[/tex]
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