Answer :
To find the derivative of the function [tex]\( t(x) = 3|x| + \frac{5}{x} \)[/tex], let's proceed step-by-step.
1. Understand the absolute value function:
The function involves an absolute value [tex]\( |x| \)[/tex]. For [tex]\( |x| \)[/tex], the derivative is defined in two different forms:
- For [tex]\( x > 0 \)[/tex], [tex]\( \frac{d}{dx} |x| = 1 \)[/tex].
- For [tex]\( x < 0 \)[/tex], [tex]\( \frac{d}{dx} |x| = -1 \)[/tex].
2. Differentiate the first term: [tex]\( 3|x| \)[/tex]
We need to consider the cases for [tex]\( x > 0 \)[/tex] and [tex]\( x < 0 \)[/tex]:
- If [tex]\( x > 0 \)[/tex], [tex]\( \frac{d}{dx} (3|x|) = 3 \cdot 1 = 3 \)[/tex].
- If [tex]\( x < 0 \)[/tex], [tex]\( \frac{d}{dx} (3|x|) = 3 \cdot (-1) = -3 \)[/tex].
To handle these cases together, we use the signum function [tex]\( \text{sign}(x) \)[/tex]:
[tex]\[ \frac{d}{dx}(3|x|) = 3 \cdot \text{sign}(x) \][/tex]
3. Differentiate the second term: [tex]\( \frac{5}{x} \)[/tex]
For [tex]\( \frac{5}{x} \)[/tex], we use the power rule:
[tex]\[ \frac{d}{dx}\left(\frac{5}{x}\right) = \frac{d}{dx}(5x^{-1}) = 5 \cdot (-1)x^{-2} = -\frac{5}{x^2} \][/tex]
4. Combine the results:
Now we combine the derivatives of each term:
[tex]\[ t'(x) = \frac{d}{dx}(3|x|) + \frac{d}{dx}\left(\frac{5}{x}\right) \][/tex]
Plugging in our earlier results:
[tex]\[ t'(x) = 3 \cdot \text{sign}(x) - \frac{5}{x^2} \][/tex]
5. Express final derivative considering real and imaginary parts:
For a more precise representation, especially considering [tex]\( x \)[/tex] might have complex components:
[tex]\[ t'(x) = 3 \left(\text{Re}(x) \frac{d}{dx} \text{Re}(x) + \text{Im}(x) \frac{d}{dx} \text{Im}(x)\right) \frac{\text{sign}(x)}{x} - \frac{5}{x^2} \][/tex]
Therefore, the final derivative of the function [tex]\( t(x) \)[/tex] is:
[tex]\[ t'(x) = 3\left(\text{Re}(x) \frac{d}{dx} \text{Re}(x) + \text{Im}(x) \frac{d}{dx} \text{Im}(x)\right) \frac{\text{sign}(x)}{x} - \frac{5}{x^2} \][/tex]
1. Understand the absolute value function:
The function involves an absolute value [tex]\( |x| \)[/tex]. For [tex]\( |x| \)[/tex], the derivative is defined in two different forms:
- For [tex]\( x > 0 \)[/tex], [tex]\( \frac{d}{dx} |x| = 1 \)[/tex].
- For [tex]\( x < 0 \)[/tex], [tex]\( \frac{d}{dx} |x| = -1 \)[/tex].
2. Differentiate the first term: [tex]\( 3|x| \)[/tex]
We need to consider the cases for [tex]\( x > 0 \)[/tex] and [tex]\( x < 0 \)[/tex]:
- If [tex]\( x > 0 \)[/tex], [tex]\( \frac{d}{dx} (3|x|) = 3 \cdot 1 = 3 \)[/tex].
- If [tex]\( x < 0 \)[/tex], [tex]\( \frac{d}{dx} (3|x|) = 3 \cdot (-1) = -3 \)[/tex].
To handle these cases together, we use the signum function [tex]\( \text{sign}(x) \)[/tex]:
[tex]\[ \frac{d}{dx}(3|x|) = 3 \cdot \text{sign}(x) \][/tex]
3. Differentiate the second term: [tex]\( \frac{5}{x} \)[/tex]
For [tex]\( \frac{5}{x} \)[/tex], we use the power rule:
[tex]\[ \frac{d}{dx}\left(\frac{5}{x}\right) = \frac{d}{dx}(5x^{-1}) = 5 \cdot (-1)x^{-2} = -\frac{5}{x^2} \][/tex]
4. Combine the results:
Now we combine the derivatives of each term:
[tex]\[ t'(x) = \frac{d}{dx}(3|x|) + \frac{d}{dx}\left(\frac{5}{x}\right) \][/tex]
Plugging in our earlier results:
[tex]\[ t'(x) = 3 \cdot \text{sign}(x) - \frac{5}{x^2} \][/tex]
5. Express final derivative considering real and imaginary parts:
For a more precise representation, especially considering [tex]\( x \)[/tex] might have complex components:
[tex]\[ t'(x) = 3 \left(\text{Re}(x) \frac{d}{dx} \text{Re}(x) + \text{Im}(x) \frac{d}{dx} \text{Im}(x)\right) \frac{\text{sign}(x)}{x} - \frac{5}{x^2} \][/tex]
Therefore, the final derivative of the function [tex]\( t(x) \)[/tex] is:
[tex]\[ t'(x) = 3\left(\text{Re}(x) \frac{d}{dx} \text{Re}(x) + \text{Im}(x) \frac{d}{dx} \text{Im}(x)\right) \frac{\text{sign}(x)}{x} - \frac{5}{x^2} \][/tex]
