Answer: The second derivative would be [tex]f''(x)=18e^{3x}+12e^{-2x}[/tex]
Step-by-step explanation:
Since we have given that
[tex]f(x)=2e^{3x}+3e^{-2x}[/tex]
We will first find the first derivative w.r.t 'x'.
As we know that
[tex]e^{mx}=me^x\\\\e^{-nx}=-ne^{-nx}[/tex]
So, it becomes,
[tex]f'(x)=6e^{3x}-6xe^{-2x}[/tex]
Now, the second derivative w.r.t 'x' would be
[tex]f''(x)=6\times 3e^{3x}-6\times -2e^{-2x}=18e^{3x}+12e^{-2x}[/tex]
Hence, the second derivative would be [tex]f''(x)=18e^{3x}+12e^{-2x}[/tex]
