# f(x) = e^x ln(x^2+3). f'(x) f"(x)?

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### 1 Answer

You need to evaluate f'(x), hence, you need to differentiate the function f(x) with respect to x , using the product rule and the chain rule, such that:

`f'(x) = (e^x)'ln(x^2+3) + e^x (ln(x^2+3))' `

`f'(x) = e^x*ln(x^2+3) + e^x*1/(x^2+3)*(x^2+3)'`

`f'(x) = e^x*ln(x^2+3) + e^x*(2x)/(x^2+3)`

You need to differeniate the function f'(x) with respect to x such that:

`f''(x) = (e^x*ln(x^2+3))' + (e^x*(2x)/(x^2+3))'`

`f''(x) = e^x*ln(x^2+3) + e^x*(2x)/(x^2+3) + ((e^x*(2x))'(x^2+3) - (e^x*(2x))(x^2+3)')/((x^2+3)^2)`

`f''(x) = e^x*ln(x^2+3) + e^x*(2x)/(x^2+3) + ((2x*e^x + 2e^x)(x^2+3) - 4x^2*e^x)/((x^2+3)^2)`

**Hence, evaluating the first and second order derivatives of the given function yields `f'(x) = e^x*ln(x^2+3) + e^x*(2x)/(x^2+3)` and `f''(x) = e^x*ln(x^2+3) + e^x*(2x)/(x^2+3) + ((2x*e^x + 2e^x)(x^2+3) - 4x^2*e^x)/((x^2+3)^2).` **