Answer :
Certainly! Let's find the third derivative of the function \( f(x) = 4x^4 + 2x^3 - 6x + 1 \) step-by-step.
### Step 1: Find the First Derivative
To find the first derivative of the function \( f(x) \), we apply the power rule of differentiation to each term:
[tex]\[ f(x) = 4x^4 + 2x^3 - 6x + 1 \][/tex]
The derivatives are:
- The derivative of \( 4x^4 \) is \( 16x^3 \)
- The derivative of \( 2x^3 \) is \( 6x^2 \)
- The derivative of \( -6x \) is \( -6 \)
- The derivative of the constant \( 1 \) is \( 0 \)
Combining these results, the first derivative \( f'(x) \) is:
[tex]\[ f'(x) = 16x^3 + 6x^2 - 6 \][/tex]
### Step 2: Find the Second Derivative
Next, we find the second derivative by differentiating \( f'(x) \):
[tex]\[ f'(x) = 16x^3 + 6x^2 - 6 \][/tex]
The derivatives are:
- The derivative of \( 16x^3 \) is \( 48x^2 \)
- The derivative of \( 6x^2 \) is \( 12x \)
- The derivative of the constant \( -6 \) is \( 0 \)
Combining these results, the second derivative \( f''(x) \) is:
[tex]\[ f''(x) = 48x^2 + 12x \][/tex]
### Step 3: Find the Third Derivative
Lastly, we find the third derivative by differentiating \( f''(x) \):
[tex]\[ f''(x) = 48x^2 + 12x \][/tex]
The derivatives are:
- The derivative of \( 48x^2 \) is \( 96x \)
- The derivative of \( 12x \) is \( 12 \)
Combining these results, the third derivative \( f'''(x) \) is:
[tex]\[ f'''(x) = 96x + 12 \][/tex]
### Final Result
Thus, the third derivative of the function \( f(x) = 4x^4 + 2x^3 - 6x + 1 \) is:
[tex]\[ f'''(x) = 96x + 12 \][/tex]
Given [tex]\( y = 12 \)[/tex], the value of the third derivative in terms of [tex]\( x \)[/tex] does not depend on [tex]\( y \)[/tex]. So our final expression remains [tex]\( 96x + 12 \)[/tex].
### Step 1: Find the First Derivative
To find the first derivative of the function \( f(x) \), we apply the power rule of differentiation to each term:
[tex]\[ f(x) = 4x^4 + 2x^3 - 6x + 1 \][/tex]
The derivatives are:
- The derivative of \( 4x^4 \) is \( 16x^3 \)
- The derivative of \( 2x^3 \) is \( 6x^2 \)
- The derivative of \( -6x \) is \( -6 \)
- The derivative of the constant \( 1 \) is \( 0 \)
Combining these results, the first derivative \( f'(x) \) is:
[tex]\[ f'(x) = 16x^3 + 6x^2 - 6 \][/tex]
### Step 2: Find the Second Derivative
Next, we find the second derivative by differentiating \( f'(x) \):
[tex]\[ f'(x) = 16x^3 + 6x^2 - 6 \][/tex]
The derivatives are:
- The derivative of \( 16x^3 \) is \( 48x^2 \)
- The derivative of \( 6x^2 \) is \( 12x \)
- The derivative of the constant \( -6 \) is \( 0 \)
Combining these results, the second derivative \( f''(x) \) is:
[tex]\[ f''(x) = 48x^2 + 12x \][/tex]
### Step 3: Find the Third Derivative
Lastly, we find the third derivative by differentiating \( f''(x) \):
[tex]\[ f''(x) = 48x^2 + 12x \][/tex]
The derivatives are:
- The derivative of \( 48x^2 \) is \( 96x \)
- The derivative of \( 12x \) is \( 12 \)
Combining these results, the third derivative \( f'''(x) \) is:
[tex]\[ f'''(x) = 96x + 12 \][/tex]
### Final Result
Thus, the third derivative of the function \( f(x) = 4x^4 + 2x^3 - 6x + 1 \) is:
[tex]\[ f'''(x) = 96x + 12 \][/tex]
Given [tex]\( y = 12 \)[/tex], the value of the third derivative in terms of [tex]\( x \)[/tex] does not depend on [tex]\( y \)[/tex]. So our final expression remains [tex]\( 96x + 12 \)[/tex].
