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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!!

S4M1R
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  • Note you could usefully write $e^x-\sin x$ as $-e^x(\sin x)$ - at the moment the natural reading of some of your lines is wrong. And then you have simply taken the factor $2$ twice right at the end, when once is all you need. – Mark Bennet Oct 10 '14 at 08:47

1 Answers1

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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 !

Traklon
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