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I don't understand line 3.

After using the Quotient Rule, how do I use the Trig identities to simplify the third line?

Also, what is the strategy when using trig identities, when do you know you are finished?

If I can choose between $\cot \ x = \frac{\cos \ x}{\sin \ x}$, which one would I choose? $\cot \ x$, or $\frac{\cos \ x}{\sin \ x}$?

Differentiate $f(x) = \dfrac{\sec x}{1+\tan x}.\;$For what values of $x$ does the graph of $f$ have a horizontal tangent?


Solution The Quotient Rule gives $$\begin{align} f'(x) & = \frac{(1+\tan x)\frac d{dx}(\sec x) - \sec x \frac d{dx}(1+\tan x)}{(1+\tan x)^2}\\ \\ & = \frac{(1+\tan x)\sec x \tan x - \sec x \cdot \sec^2 x}{(1+\tan x)^2}\\\\ &= \frac{\sec x(\tan x + \tan^2 x - \sec^2 x)}{(1+\tan s)^2}\\ \\ &= \frac{\sec x(\tan x - 1)}{(1 + \tan x)^2} \end{align}$$

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  • $\begingroup$ They just factor out a $\sec x$ from the terms on top. $\endgroup$ – Mnifldz Mar 13 '15 at 14:50
  • $\begingroup$ You're finished with calc part of the problem by the end of line $2$. $\endgroup$ – ganeshie8 Mar 13 '15 at 14:50
  • $\begingroup$ Mnifldz when you said they factored out the secant. You can remove a secant from a sum of products? What law of real numbers is this? $\endgroup$ – Cetshwayo Mar 13 '15 at 14:57
  • $\begingroup$ To you point about simplification - the last expression is equal to $\frac{\sin x-\cos x}{(\sin x+\cos x)^2}$, which some would argue is more elegant. $\endgroup$ – Mark Viola Mar 13 '15 at 15:28
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To answer one of your querry:

Factor out $sec\ x$ and multiply $(1+tan\ x)tan\ x$ in the numerator of line 2. We now have the third line. Using the identity $tan^2+1=sec^2x$ we have line 4.

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  • $\begingroup$ So you are always aiming for the identity to be very simple, as in one identity. For example cot x, instead of cos x / sin x? $\endgroup$ – Cetshwayo Mar 13 '15 at 15:00
  • $\begingroup$ I would say yes for simplicity. Also based from my experience if the original function involves say sec and tan, the answer is always in terms of sec and tan OR sec or tan. Hope this helps. $\endgroup$ – Jr Antalan Mar 13 '15 at 15:03
  • $\begingroup$ I guess this is more of an Algebra question; however, line #2, when factoring out the sec x from the second terms, how should I view it? Since this is the inverse of the distributive property a(b+c)=ab+ac. Yet, I don't see that with this question. $\endgroup$ – Cetshwayo Mar 13 '15 at 15:09
  • $\begingroup$ Again based on my experience, most of the time, factoring out a common factor helps a lot in differentiation :-). $\endgroup$ – Jr Antalan Mar 13 '15 at 15:22
  • $\begingroup$ How did this answer either of the OP's questions? The OP asked about simplifying Line 3, not arriving at it from Line 2. And the OP wanted some guidelines as to knowing when to "stop" simplifying. $\endgroup$ – Mark Viola Mar 13 '15 at 15:33
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Question 1: Third Line Issue $\tan ^2x - \sec ^2 x = \left(\frac{\sin x}{\cos x}\right)^2-\left(\frac{1}{\cos x}\right)^2=\frac{\sin ^2x-1}{\cos ^2x}=\frac{-\cos ^2x}{\cos ^2x}=-1$

Question 2: General Question It is a matter of choice really. Inasmuch as you are talking about equivalent expressions, then these expression are equally valid. That said, one often chooses the form that is more "concise," whatever "concise" means to you.

So, in the case you mentioned, $\cot x$ versus $\frac{\cos x}{\sin x}$, which seems more concise to you? Moreover, the final expression is equivalent to $\frac{\sin x-\cos x}{\left(\sin x+\cos x\right)^2}$. Does that appear more elegant than Line 4 of the post?

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  • $\begingroup$ tan x would seem more concise $\endgroup$ – Cetshwayo Mar 13 '15 at 15:20
  • $\begingroup$ That sounds very reasonable to me! I do understand your question and hope that I've helped some. $\endgroup$ – Mark Viola Mar 13 '15 at 15:23
  • $\begingroup$ To you point about simplification - the last expression is equal to $\frac{\sin x-\cos x}{(\sin x+\cos x)^2}$, which some would argue is more elegant. $\endgroup$ – Mark Viola Mar 13 '15 at 15:29

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