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I have the following equation to calculate its derivative:

$$f(x) = e^{3x}(\sin(x)+\cos(x))$$

I used the product rule and I got this answer:

$$e^{3x}(\sin(x)+\cos(x))+e^{3x}(\cos(x)-\sin(x))$$

But the answer at the end of the book is:

$$3e^{3x}(\sin(x)+\cos(x))+e^{3x}(\cos(x)-\sin(x))$$

I am scratching my head to find where that $3$ came from. I think I have to use chain rule here as well, but I am not sure to take which part of equation as $z$.

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    $\begingroup$ The derivative of $e^{3x}$ is $3e^{3x}$. $\endgroup$ – azarel Apr 12 '12 at 18:31
  • $\begingroup$ You could also simplify the answer further by factoring out $e^{3x}$ and combining like terms. $\endgroup$ – Mike Apr 12 '12 at 20:06
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Let $g(x)=e^{3x}$ and $h(x)=\sin(x)+\cos(x)$.Then $f(x)=g(x)\cdot h(x)$. The product rule states $f'(x)=g'(x)h(x)+g(x)h'(x)$.In our case $$(e^{3x})'(\sin(x)+\cos(x))+(e^{3x})(\sin(x)+\cos(x))'=3e^{3x}(\sin(x)+\cos(x))+e^{3x}(\cos(x)-\sin(x))$$

Watch out that $ (e^{3x})'=3e^{3x}$. That's because $e^{3x}=e^{z}$ where $z=3x$. Therefore $\frac{de^z}{dx}=\frac{de^z}{dz}\frac{dz}{dx}=e^z\cdot z'(x)$. Substituting back, we got: $\frac{de^{3x}}{dx}=e^{3x}\cdot (3x)'=3e^{3x}$

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The 3 comes from the derivative of $e^{3x}$. The derivative of $e^{3x}$ is $3e^{3x}$.

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