First step, derivative of the constant:
[tex]f^{\prime}(x)=\frac{1}{2}\frac{d(e^{-3x^2+tan(3x+5)})}{dx}[/tex]
Second step, the chain rule:
[tex]\begin{gathered} =\frac{1}{2}e^{-3x^2+tan(3x+5)}*\frac{d(-3x^2+tan(3x+5))}{dx} \\ =\frac{1}{2}e^{-3x^2+tan(3x+5)}*\frac{(-6x+sec^2(3x+5)*3)}{} \end{gathered}[/tex]
Where we applied that the derivative of tangent is the secant^2.
Hence the answer is:
[tex]\frac{1}{2}e^{-3x^2+tan(3x+5)}*\frac{(-6x+3sec^2(3x+5))}{}[/tex]
