Answer :
To find the derivative [tex]\( f'(x) \)[/tex] and evaluate it at [tex]\( x = 5 \)[/tex] for the function [tex]\( f(x) = 3 + \frac{4}{x} + \frac{4}{x^2} \)[/tex], follow these steps:
### Step-by-Step Solution:
1. Identify the function:
[tex]\[ f(x) = 3 + \frac{4}{x} + \frac{4}{x^2} \][/tex]
2. Differentiate the function:
We differentiate each term of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
- The derivative of [tex]\( 3 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 0 \)[/tex].
- The derivative of [tex]\( \frac{4}{x} \)[/tex]:
Recall that [tex]\( \frac{4}{x} = 4x^{-1} \)[/tex], and using the power rule:
[tex]\[ \frac{d}{dx} (4x^{-1}) = 4 \cdot (-1)x^{-2} = -\frac{4}{x^2} \][/tex]
- The derivative of [tex]\( \frac{4}{x^2} \)[/tex]:
Recall that [tex]\( \frac{4}{x^2} = 4x^{-2} \)[/tex], and using the power rule:
[tex]\[ \frac{d}{dx} (4x^{-2}) = 4 \cdot (-2)x^{-3} = -\frac{8}{x^3} \][/tex]
Combining these results gives:
[tex]\[ f'(x) = 0 - \frac{4}{x^2} - \frac{8}{x^3} = -\frac{4}{x^2} - \frac{8}{x^3} \][/tex]
3. Evaluate [tex]\( f'(x) \)[/tex] at [tex]\( x = 5 \)[/tex]:
Substitute [tex]\( x = 5 \)[/tex] into the derivative [tex]\( f'(x) = -\frac{4}{x^2} - \frac{8}{x^3} \)[/tex]:
[tex]\[ f'(5) = -\frac{4}{5^2} - \frac{8}{5^3} = -\frac{4}{25} - \frac{8}{125} \][/tex]
Find a common denominator to combine the fractions:
[tex]\[ -\frac{4}{25} = -\frac{20}{125} \][/tex]
So:
[tex]\[ f'(5) = -\frac{20}{125} - \frac{8}{125} = -\frac{28}{125} \][/tex]
### Summary:
- The first derivative of [tex]\( f(x) = 3 + \frac{4}{x} + \frac{4}{x^2} \)[/tex] is:
[tex]\[ f'(x) = -\frac{4}{x^2} - \frac{8}{x^3} \][/tex]
- The value of the derivative at [tex]\( x = 5 \)[/tex] is:
[tex]\[ f'(5) = -\frac{28}{125} \][/tex]
### Step-by-Step Solution:
1. Identify the function:
[tex]\[ f(x) = 3 + \frac{4}{x} + \frac{4}{x^2} \][/tex]
2. Differentiate the function:
We differentiate each term of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
- The derivative of [tex]\( 3 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 0 \)[/tex].
- The derivative of [tex]\( \frac{4}{x} \)[/tex]:
Recall that [tex]\( \frac{4}{x} = 4x^{-1} \)[/tex], and using the power rule:
[tex]\[ \frac{d}{dx} (4x^{-1}) = 4 \cdot (-1)x^{-2} = -\frac{4}{x^2} \][/tex]
- The derivative of [tex]\( \frac{4}{x^2} \)[/tex]:
Recall that [tex]\( \frac{4}{x^2} = 4x^{-2} \)[/tex], and using the power rule:
[tex]\[ \frac{d}{dx} (4x^{-2}) = 4 \cdot (-2)x^{-3} = -\frac{8}{x^3} \][/tex]
Combining these results gives:
[tex]\[ f'(x) = 0 - \frac{4}{x^2} - \frac{8}{x^3} = -\frac{4}{x^2} - \frac{8}{x^3} \][/tex]
3. Evaluate [tex]\( f'(x) \)[/tex] at [tex]\( x = 5 \)[/tex]:
Substitute [tex]\( x = 5 \)[/tex] into the derivative [tex]\( f'(x) = -\frac{4}{x^2} - \frac{8}{x^3} \)[/tex]:
[tex]\[ f'(5) = -\frac{4}{5^2} - \frac{8}{5^3} = -\frac{4}{25} - \frac{8}{125} \][/tex]
Find a common denominator to combine the fractions:
[tex]\[ -\frac{4}{25} = -\frac{20}{125} \][/tex]
So:
[tex]\[ f'(5) = -\frac{20}{125} - \frac{8}{125} = -\frac{28}{125} \][/tex]
### Summary:
- The first derivative of [tex]\( f(x) = 3 + \frac{4}{x} + \frac{4}{x^2} \)[/tex] is:
[tex]\[ f'(x) = -\frac{4}{x^2} - \frac{8}{x^3} \][/tex]
- The value of the derivative at [tex]\( x = 5 \)[/tex] is:
[tex]\[ f'(5) = -\frac{28}{125} \][/tex]
