To find the derivative of the function [tex]\( f(x) = \sqrt{5x^2 + 5x + 8} \)[/tex] and then evaluate the derivative at [tex]\( x = 3 \)[/tex], we can proceed step-by-step as follows:
### Step 1: Define the function
We have the function:
[tex]\[ f(x) = \sqrt{5x^2 + 5x + 8} \][/tex]
### Step 2: Apply the chain rule
To differentiate [tex]\( f(x) \)[/tex], we use the chain rule. Let [tex]\( u = 5x^2 + 5x + 8 \)[/tex], so that [tex]\( f(x) = \sqrt{u} \)[/tex].
The chain rule states:
[tex]\[ \frac{d}{dx} f(x) = \frac{d}{dx} \sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \][/tex]
### Step 3: Differentiate the inner function [tex]\( u \)[/tex]
Next, we differentiate [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ u = 5x^2 + 5x + 8 \][/tex]
[tex]\[ \frac{du}{dx} = \frac{d}{dx}(5x^2) + \frac{d}{dx}(5x) + \frac{d}{dx}(8) \][/tex]
[tex]\[ \frac{du}{dx} = 10x + 5 + 0 = 10x + 5 \][/tex]
### Step 4: Combine results from chain rule
Putting it all together, we get:
[tex]\[ f'(x) = \frac{1}{2\sqrt{5x^2 + 5x + 8}} \cdot (10x + 5) \][/tex]
[tex]\[ f'(x) = \frac{10x + 5}{2\sqrt{5x^2 + 5x + 8}} \][/tex]
[tex]\[ f'(x) = \frac{5(2x + 1)}{2\sqrt{5x^2 + 5x + 8}} \][/tex]
Therefore, the derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{5x + \frac{5}{2}}{\sqrt{5x^2 + 5x + 8}} \][/tex]
### Step 5: Evaluate the derivative at [tex]\( x = 3 \)[/tex]
Now, we substitute [tex]\( x = 3 \)[/tex] into the derivative:
[tex]\[ f'(3) = \frac{5(3) + \frac{5}{2}}{\sqrt{5(3)^2 + 5(3) + 8}} \][/tex]
[tex]\[ f'(3) = \frac{15 + \frac{5}{2}}{\sqrt{45 + 15 + 8}} \][/tex]
[tex]\[ f'(3) = \frac{15 + 2.5}{\sqrt{68}} \][/tex]
[tex]\[ f'(3) = \frac{17.5}{\sqrt{68}} \][/tex]
To make the fraction simpler, we get:
[tex]\[ f'(3) = \frac{35}{2\sqrt{68}} = \frac{35\sqrt{68}}{2 \cdot 68} = \frac{35\sqrt{17}}{68} \][/tex]
Thus, the value of the derivative at [tex]\( x = 3 \)[/tex] is:
[tex]\[ f'(3) = \frac{35\sqrt{17}}{68} \][/tex]
In summary:
[tex]\[ f'(x) = \frac{5x + \frac{5}{2}}{\sqrt{5x^2 + 5x + 8}} \][/tex]
[tex]\[ f'(3) = \frac{35\sqrt{17}}{68} \][/tex]
