Derivative of constant divided by equation I'm having trouble figuring out how to derive $\frac{5}{4 + 3\cos{2x}}$.
Using the $D\frac{f}{g} = \frac{gDf - fDg}{g^2}$ rule doesn't seem to work: it results in zero.
 A: Hint: $f Dg = 5 (-6 \sin2x)$
Do you see your issue now?
A: Well,
With time you'll get practice with derivatives and you'll do the derivatives of $f$ and $g$ mentally to put in there. However, to get started write down the derivatives. You have:
$$f(x) = 5$$
$$g(x) = 4+ 3\cos2x$$
$$h(x)=\left(\frac{f}{g}\right)(x)$$
Hence you have $f'(x) = 0$ and $g'(x) = -6 \sin2x$. Now you plug them into the equation that gives the derivative of $h$:
$$h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$
Substituting you get:
$$h'(x)=\frac{30\sin2x}{(4+3\cos2x)^2}$$
Which is not zero. Do it to get started, write down the functions involved, the derivatives and work out what you want. With time you get practive and mentally you're goint to think : "well, derivative of $f$ is this, derivative of $g$ is that now let's put them in the formula".
A: Actually, you can simply use
$$
\left(\frac{5}{f}\right)'=-\frac{5f'}{f^2}
$$
with $f(x)=4+3\cos(2x)$.
A: For this case,
it is handy to know that
$(1/f)' = -f'/f^2$.
The 5 in the numerator is a constant,
so it stays.
Setting $f(x) = 4+3\cos(2x)$,
$f'(x) = 6\sin(2x)$,
so $(1/f)' = -6\sin(2x)/(4+3\cos(2x))^2$
and the final result is
5 times this.
