Take the 2-minute tour ×  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:   But the answer at
the end of the book is:   I am scratching my head to find where
that $3$ came from. I think I have to use chaing rule here as
well, but I am not sure to take which part of equation as $z$.
share|improve this question The derivative of $e^{3x}$ is
$3e^{3x}$. –  azarel Apr 12 '12 at 18:31 You could also simplify
the answer further by factoring out $e^{3x}$ and combining like
terms. –  Mike Apr 12 '12 at 20:06  1 Answer 1  up vote 2 down
vote accepted  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