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Sure! Let's start by finding the derivative [tex]\( f'(x) \)[/tex] of the function [tex]\( f(x) = \sqrt{4x^2 + 3x + 8} \)[/tex], and then evaluate [tex]\( f'(3) \)[/tex].
### Step-by-step Solution
#### Step 1: Differentiate [tex]\( f(x) = \sqrt{4x^2 + 3x + 8} \)[/tex]
To find [tex]\( f'(x) \)[/tex], we'll use the chain rule. Recall that for a function of the form [tex]\( f(x) = \sqrt{g(x)} \)[/tex], the derivative [tex]\( f'(x) \)[/tex] can be found using:
[tex]\[ \frac{d}{dx} \left( \sqrt{g(x)} \right) = \frac{1}{2\sqrt{g(x)}} \cdot g'(x) \][/tex]
Here, [tex]\( g(x) = 4x^2 + 3x + 8 \)[/tex].
First, find [tex]\( g'(x) \)[/tex]:
[tex]\[ g(x) = 4x^2 + 3x + 8 \][/tex]
[tex]\[ g'(x) = \frac{d}{dx} (4x^2) + \frac{d}{dx} (3x) + \frac{d}{dx} (8) \][/tex]
[tex]\[ g'(x) = 8x + 3 \][/tex]
Now, applying the chain rule:
[tex]\[ f'(x) = \frac{1}{2\sqrt{4x^2 + 3x + 8}} \cdot (8x + 3) \][/tex]
This simplifies to:
[tex]\[ f'(x) = \frac{8x + 3}{2\sqrt{4x^2 + 3x + 8}} \][/tex]
[tex]\[ f'(x) = \frac{4x + \frac{3}{2}}{\sqrt{4x^2 + 3x + 8}} \][/tex]
#### Step 2: Evaluate [tex]\( f'(3) \)[/tex]
Next, we need to evaluate the derivative at [tex]\( x = 3 \)[/tex]:
[tex]\[ f'(3) = \frac{4(3) + \frac{3}{2}}{\sqrt{4(3)^2 + 3(3) + 8}} \][/tex]
Calculate the numerator:
[tex]\[ 4(3) + \frac{3}{2} = 12 + \frac{3}{2} = 12 + 1.5 = 13.5 \][/tex]
Calculate the denominator:
[tex]\[ \sqrt{4(3)^2 + 3(3) + 8} = \sqrt{4 \cdot 9 + 9 + 8} = \sqrt{36 + 9 + 8} = \sqrt{53} \][/tex]
Thus:
[tex]\[ f'(3) = \frac{13.5}{\sqrt{53}} \][/tex]
To rationalize the denominator:
[tex]\[ f'(3) = \frac{13.5}{\sqrt{53}} \cdot \frac{\sqrt{53}}{\sqrt{53}} = \frac{13.5 \sqrt{53}}{53} \][/tex]
Convert 13.5 to a fraction:
[tex]\[ 13.5 = \frac{27}{2} \][/tex]
So:
[tex]\[ f'(3) = \frac{\frac{27}{2} \sqrt{53}}{53} = \frac{27 \sqrt{53}}{2 \cdot 53} = \frac{27 \sqrt{53}}{106} \][/tex]
#### Summary
- The derivative [tex]\( f'(x) \)[/tex] is given by:
[tex]\[ f'(x) = \frac{4x + \frac{3}{2}}{\sqrt{4x^2 + 3x + 8}} \][/tex]
- The value [tex]\( f'(3) \)[/tex] is:
[tex]\[ f'(3) = \frac{27 \sqrt{53}}{106} \][/tex]
Thus,
[tex]\[ f'(x) = \left( \frac{4x + \frac{3}{2}}{\sqrt{4x^2 + 3x + 8}} \right) \][/tex]
[tex]\[ f'(3) = \left( \frac{27 \sqrt{53}}{106} \right) \][/tex]
### Step-by-step Solution
#### Step 1: Differentiate [tex]\( f(x) = \sqrt{4x^2 + 3x + 8} \)[/tex]
To find [tex]\( f'(x) \)[/tex], we'll use the chain rule. Recall that for a function of the form [tex]\( f(x) = \sqrt{g(x)} \)[/tex], the derivative [tex]\( f'(x) \)[/tex] can be found using:
[tex]\[ \frac{d}{dx} \left( \sqrt{g(x)} \right) = \frac{1}{2\sqrt{g(x)}} \cdot g'(x) \][/tex]
Here, [tex]\( g(x) = 4x^2 + 3x + 8 \)[/tex].
First, find [tex]\( g'(x) \)[/tex]:
[tex]\[ g(x) = 4x^2 + 3x + 8 \][/tex]
[tex]\[ g'(x) = \frac{d}{dx} (4x^2) + \frac{d}{dx} (3x) + \frac{d}{dx} (8) \][/tex]
[tex]\[ g'(x) = 8x + 3 \][/tex]
Now, applying the chain rule:
[tex]\[ f'(x) = \frac{1}{2\sqrt{4x^2 + 3x + 8}} \cdot (8x + 3) \][/tex]
This simplifies to:
[tex]\[ f'(x) = \frac{8x + 3}{2\sqrt{4x^2 + 3x + 8}} \][/tex]
[tex]\[ f'(x) = \frac{4x + \frac{3}{2}}{\sqrt{4x^2 + 3x + 8}} \][/tex]
#### Step 2: Evaluate [tex]\( f'(3) \)[/tex]
Next, we need to evaluate the derivative at [tex]\( x = 3 \)[/tex]:
[tex]\[ f'(3) = \frac{4(3) + \frac{3}{2}}{\sqrt{4(3)^2 + 3(3) + 8}} \][/tex]
Calculate the numerator:
[tex]\[ 4(3) + \frac{3}{2} = 12 + \frac{3}{2} = 12 + 1.5 = 13.5 \][/tex]
Calculate the denominator:
[tex]\[ \sqrt{4(3)^2 + 3(3) + 8} = \sqrt{4 \cdot 9 + 9 + 8} = \sqrt{36 + 9 + 8} = \sqrt{53} \][/tex]
Thus:
[tex]\[ f'(3) = \frac{13.5}{\sqrt{53}} \][/tex]
To rationalize the denominator:
[tex]\[ f'(3) = \frac{13.5}{\sqrt{53}} \cdot \frac{\sqrt{53}}{\sqrt{53}} = \frac{13.5 \sqrt{53}}{53} \][/tex]
Convert 13.5 to a fraction:
[tex]\[ 13.5 = \frac{27}{2} \][/tex]
So:
[tex]\[ f'(3) = \frac{\frac{27}{2} \sqrt{53}}{53} = \frac{27 \sqrt{53}}{2 \cdot 53} = \frac{27 \sqrt{53}}{106} \][/tex]
#### Summary
- The derivative [tex]\( f'(x) \)[/tex] is given by:
[tex]\[ f'(x) = \frac{4x + \frac{3}{2}}{\sqrt{4x^2 + 3x + 8}} \][/tex]
- The value [tex]\( f'(3) \)[/tex] is:
[tex]\[ f'(3) = \frac{27 \sqrt{53}}{106} \][/tex]
Thus,
[tex]\[ f'(x) = \left( \frac{4x + \frac{3}{2}}{\sqrt{4x^2 + 3x + 8}} \right) \][/tex]
[tex]\[ f'(3) = \left( \frac{27 \sqrt{53}}{106} \right) \][/tex]
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