Higher order derivatives are not very difficult. The idea is that, after taking the first derivative, you can take the derivative again to get the second derivative, and so on. Here is a quick example.
Example   Find the second derivative of \(f(x)=3x^5 + 2x+1\).
The first derivative is \( f'(x) = 15x^4 + 2\).
Taking the derivative again, gives us the second derivative \( f''(x) = 60x^3 \).
Notation   We can continue taking as many derivatives as we want. The third derivative is written \( f'''(x) \) or \( f^{(3)}(x)\). Notice that the derivative number is written in parentheses so that we can tell we are talking about a derivative and not a power, which we would write \( f^{3}(x)\). Starting with the fourth derivative, we abandon the 'multiple quote' notation and write \(f^{(4)}(x) \) and so on.
Okay, let's work some practice problems calculating higher order derivatives. As before, do not use the limit definition to calculate the derivative and give your answers in completely factored simplified form.
Problem Statement 

Calculate \(f''(x)\) of \(f(x)=3x^4+3x^2+2x \)
Final Answer 

Problem Statement 

Calculate \(f''(x)\) of \(f(x)=3x^4+3x^2+2x \)
Solution 

The first derivative is \(\displaystyle{\frac{d}{dx}[3x^4+3x^2+2x]=12x^3+6x+2}\)
To find the second derivative, we take the derivative of the first derivative.
\(\displaystyle{\frac{d}{dx}[12x^3+6x+2]=36x^2+6}\)
Final Answer 

\(\displaystyle{6(6x^2+1)}\) 
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Okay, you are ready to move on. If you need a suggestion on where to go next, the product rule is the logical next step.
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