how to solve $f''(e^{x} \cdot \sin x)$ How do I find the derivative $f''(e^{x} \cdot \sin x)$.
I start to find $f'(x)$ by using the product rule $f'(e^{x} \cdot \sin x) = e^{x} \cdot \sin x + e^{x} \cdot \cos x = e^{x}(\sin x + \cos x)$
Now when I have $f'(x)$ I use it to find $f''(x)$.
I split upp the problem by finding the derivative of $e^{x} \cdot \sin x$ first and then the derivative of $e^{x} \cdot \cos x$
$f'(e^{x} \cdot \sin x) = e^{x} \sin x + e^{x} \cos x$
$f'(e^{x} \cdot \cos x) = e^{x} - \sin x + e^{x} \cos x$
Then I plug this values back to $f''(x)$
$f''(x) = (e^{x} \sin x + e^{x} \cos x) + (e^{x} - \sin x + e^{x} \cos x)$
$f''(x) = 2 e^{x} (2 \cos x)$
I know this is wrong answer, the correct answer is $f''(x) = 2 e^{x} (\cos x)$
Where did it go wrong in my calculation?
Thanks!!

The last line is wrong : 
$f''(x) = (e^{x} \sin x + e^{x} \cos x) + (e^{x} *(-\sin x) + e^{x} \cos x)$
$f''(x) = 2 e^{x} (\cos x)$
The $2 \cos x$ just came out of inattention !
