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I'm having a very difficult time understanding how find the derivative of trig functions. Here's one that I've been trying to crack for the half hour: $$y = \frac{11}{\sin x} + \frac{1}{\cot x}$$

I know that $$\frac{d}{dx}(\sin x) = \cos x$$ and that $$\frac{d}{dx}(\cot x) = -\csc^2 x$$

I also know that the final answer is $$y' = -11\csc x\cot x+\sec^2 x$$

But no matter what I try I can't seem to come up with that answer. What I really need is systematic or step by step approcah that I could apply to these types of problems, no matter what format the're in. Because there's many more like this, except in different formats, that I have to solve.

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up vote 5 down vote accepted


$$y = \frac{11}{\sin x} + \frac{1}{\cot x} = 11\csc x + \tan x$$

Now just compute $\;y' = \dfrac{d}{dx}(11\csc x +\tan x)$

As noted in the comment below: When dealing with more complicated fractions which involve, e.g., sums or differences in the numerator and/or denominator, you can do two things:

  • check to see if either numerator or denominator can be expressed in an equivalent form, based on trigonometric identities (for a simple example: $1 - \sin^2x = \cos^2x$;

  • use the quotient rule, which you'll want to have in your "tool-box": $$\displaystyle\left(\frac {f(x)}{ g(x)}\right)' = \frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}$$

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That makes perfect sense now for that problem, but I'm having trouble applying that same approach to this problem $$p = \frac{6+secq}{6-secq}$$ I tried moving the bottom -sec(q) to the top and making it -cos(q) but am not able to get even close to the correct derivative of that function. – Scott Feb 12 '13 at 2:33
Can you try the quotient rule?: $\displaystyle\left(\frac {f(q)}{ g(q)}\right)' = \frac{f'(q)g(q)-g'(q)f(q)}{g(q)^2}$, where $f(q) = 6 + \sec q$, and $g(q) = 6 - \sec q$? – amWhy Feb 12 '13 at 2:38
These are starting to click now. – Scott Feb 12 '13 at 2:41

The quotient rule says $\displaystyle\left(\frac fg\right)' = \frac{f'g-g'f}{g^2}$, and its special case, the reciprocal rule says $\displaystyle\left(\frac1g\right)' = \frac{-g'}{g^2}$.

You can apply the reciprocal rule to $\dfrac{1}{\sin x}$ and the quotient rule to $\dfrac{1}{\cot x} = \tan x=\dfrac{\sin x}{\cos x}$.

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