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I'm posting my initial work up to this point. Criticism welcomed!


Using the formula $(fgh)' = f'gh+fg'h + fgh'$, differentiate$$y = 9x^2\sin x \tan x$$

$$\begin{align} y' &= 9\frac d{dx}(x^2)\sin x \tan x + 9x^2 \frac d{dx} (\sin x) \tan x + 9x^2\sin x \frac d{dx}(\tan x)\\ \\ & = 9(2x\sin x \tan x + 9x^2-\cos x \tan x + 9x^2\sin x \sec^2x\\ \\ &=9\Big(2x\sin x \tan x + x^2 -\cos x \tan x + x^2\sin x \sec^2 x\Big) \end{align}$$


Have I done it correctly up and until this point?

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  • $\begingroup$ Do you already know about logarithmic differentiation ? If yes, this could simplify life. If you want, I could elaborate. Just post. $\endgroup$ – Claude Leibovici Mar 14 '15 at 6:34
  • $\begingroup$ Cetshwayo: I commend you for your perseverance and the effort you always show in your posts. I can see the progress you've made and continue to make! $\endgroup$ – Namaste Mar 14 '15 at 13:09
  • $\begingroup$ @amWhy Thanks so much! $\endgroup$ – Cetshwayo Mar 15 '15 at 0:51
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\begin{aligned} \frac{d}{dx}(9x^2 \sin(x) \tan(x)) & = \frac{d}{dx}[9x^2]\sin(x) \tan(x) + 9x^2 \frac{d}{dx}[\sin(x)]\tan(x)+9x^2 \sin(x) \frac{d}{dx}[\tan(x)] & \\ & = 9(2x) \sin(x) \tan(x)+9x^2 \cos(x) \tan(x) + 9x^2 \sin(x) \sec^2(x)& \end{aligned}

and $\cos(x)\tan(x) = \sin(x)$, so:

\begin{aligned} \frac{d}{dx}(9x^2 \sin(x) \tan(x)) & = \color{red}{9(2x) \sin(x) \tan(x)+9x^2 \cos(x) \tan(x) + 9x^2 \sin(x) \sec^2(x)} & \\ & = 9(2x) \sin(x) \tan(x)+9x^2 \sin(x) + 9x^2 \sin(x) \sec^2(x) & \\ & = 9x\sin(x)[2\tan(x) + x + x\sec^2(x)]& \end{aligned}

So the derivative is:

$$\boxed{9x\sin(x)[2\tan(x) + x + x\sec^2(x)]}$$

I noticed these errors in your work: In the $\color{red}{\mathrm{red}}$ line, you wrote a minus sign. I suspect you might have thought that $\frac{d}{dx}[\sin(x)] = -\cos(x)$. In your final line, you probably misinterpreted that minus sign no longer as a negative, but a difference:

$$9[2x \sin(x) \tan(x) + \boxed{x^2 -\cos(x) \tan(x)} + x^2 \sin(x) \sec^2(x)],$$ the $\boxed{\mathrm{boxed}}$ part should actually be

$$x^2 \cos(x)\tan(x).$$

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    $\begingroup$ What happened to the nine for the derivative? $\endgroup$ – Cetshwayo Mar 13 '15 at 23:30
  • $\begingroup$ Corrected. Thanks! $\endgroup$ – Sultan of Swing Mar 13 '15 at 23:31
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    $\begingroup$ I also added to the explanation of your mistakes too, which is something to look out for in the future! $\endgroup$ – Sultan of Swing Mar 13 '15 at 23:36
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Your notation $$ 9x^2 \dfrac{d}{dx}[\sin x] \tan x $$ is a bit confusing. It's better to write $$ 9x^2 \tan x \dfrac{d}{dx}[\sin x] $$ and here is your mistake:

$$ 9x^2 \tan x \dfrac{d}{dx}[\sin x] =9x^2\tan x \cos x $$ So: the minus sign is wrong and $9x^2$ is multiplied not summed.

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  • $\begingroup$ I am following the formula. Do you suggest another strategy? What you wrote lacked some key information for me to understand what you meant. $\endgroup$ – Cetshwayo Mar 13 '15 at 23:15
  • $\begingroup$ You have followed the formula in the first step, but in the second you have transformed a product in a sum. Maybe this is because you have introduced a minus sign (wrong) without using parentesis. $\endgroup$ – Emilio Novati Mar 13 '15 at 23:19
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You have a parenthesis error

Line 3 should read $$ 9(2x)\sin x \tan x +9x^2(-\cos x)\tan x + 9x^2 \sin x \sec^2 x $$ You can actually factor out $9x\sin x$ because $\cos x \tan x = \sin x$.

Continuing $$ 9x \sin x (2\tan x - x + x \sec^2 x) \\ 9x \sin x (2\tan x +x (\sec^2 x - 1) )\\ $$

Then you can use some trig identities to simplify further.

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  • $\begingroup$ Can you elaborate what do you mean when you typed I made a parenthesis error? Also, what are the rules when it comes to factoring and expanding polynomials such as this? I treat them for the most part as un multiplied functions. Am I looking at them wrong? What is the best strategy to address these types of problems. $\endgroup$ – Cetshwayo Mar 13 '15 at 23:13
  • $\begingroup$ The way you had it written, it looked like $9x^2$ minus $\cos x \tan x$, instead of $9 x^2$ times $- \cos x \tan x$. Looking over again. The problem doesn't explicitly say to make the results "look nice". I saw a $\sec^2 x$ and a $1$. I know that equals $\csc^2 x$. It's just a pattern I saw because I've been doing this for awhile. $\endgroup$ – maxbaroi Mar 13 '15 at 23:15
  • $\begingroup$ Well isn't that what it's supposed to be? Is't the entire expression considered negative? $\endgroup$ – Cetshwayo Mar 13 '15 at 23:17
  • $\begingroup$ That's not what it's supposed to be. There is a big difference between $9x^2 (-\cos x)\tan x$ and $9x^2 - \cos x \tan x$. Pick $101 \pi$. The first equation is 0 at that point, while the second equation is near a million. $\endgroup$ – maxbaroi Mar 13 '15 at 23:33

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