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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.

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  • $\begingroup$ Hint: When you take the derivative of the numerator it is zero, but when you take the derivative of the denominator times the constant divided by the square, it is not zero. What do you get? $\endgroup$
    – Amzoti
    Mar 17, 2013 at 23:12
  • $\begingroup$ Note: an equation is something with an equals sign in it. $4+3\cos2x$ is not an equation. $\endgroup$ Mar 17, 2013 at 23:19
  • $\begingroup$ Or: you could skip the whole numerator/denominator thing, move the 5 to in front of the D, and take the derivative of $(4+3cos2x)^{-1}$ $\endgroup$
    – DJohnM
    Mar 17, 2013 at 23:20

4 Answers 4

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Hint: $f Dg = 5 (-6 \sin2x)$

Do you see your issue now?

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  • $\begingroup$ Well-done, if even only for future reference (the OP didn't seem to respond anywhere). $\endgroup$
    – amWhy
    Apr 19, 2013 at 0:33
  • $\begingroup$ @amWhy: yeah - those are always tough for me, just to know they got it, but like you said - oh well! thx $\endgroup$
    – Amzoti
    Apr 19, 2013 at 0:40
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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".

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Actually, you can simply use $$ \left(\frac{5}{f}\right)'=-\frac{5f'}{f^2} $$ with $f(x)=4+3\cos(2x)$.

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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.

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