# Determine the first and second derivative of the function f(x)=3xe^xf(x)=3xe^x f'(x)= 3xe^x + 3e^x and f''(x)= 3xe^x+6e^x Is this correct? Explain please.

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### 1 Answer

You need to differentiate the function with respect to x using the product rule such that:

`f'(x) = (3x)'*(e^x) + (3x)*(e^x)'`

You should notice that the function f(x) is a product of two functions, monomial 3x and exponential `e^x` .

`f'(x) = 3*(e^x) + (3x)*(e^x)`

You need to find the second derivative, hence you need to differentiate the function f'(x) with respect to x such that:

`f"(x) = 3e^x + ((3x)'*(e^x) + (3x)*(e^x)')`

Notice that you need to use again the product rule for the term `(3x)*(e^x).`

`f"(x) = 3e^x + 3e^x + (3x)*(e^x)`

`f"(x) = 6e^x + (3x)*(e^x)`

**Hence, evaluating the first derivative and the second derivative yields `f'(x) = 3*(e^x) + (3x)*(e^x) ; f"(x) = 6e^x + (3x)*(e^x).` **