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 March 25th, 2017, 12:41 PM #1 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 Derivative with step/answer; can't figure why? Find derivative of sin^2(tan2x) 2sin(tan2x)cos(tan2x)sec^2(2x)(2) =4sec^2(2x)sin(tan2x)cos(tan2x) First, are both the provided middle step and answer correct? If so... --------------------------------- I've been over this problem but can't get to that middle step. Is the first part using the combination of Power Rule and Chain Rule? d/dx [sin^2(tan2x)] = 2sin(cos) du/dx [sin^2(tan2x)] how then to get to the middle step? Last edited by Seventy7; March 25th, 2017 at 12:57 PM. March 25th, 2017, 01:10 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 $y(x) = \sin^2(\tan(2x))$ can look at this as 3 functions $f(x) = \tan(2x), ~\dfrac{df}{dx}=2\sec^2(x)$ $g(x) = \sin(x),~\dfrac{dg}{dx}=\cos(x)$ $h(x) = x^2,~\dfrac{dh}{dx}=2x$ $y(x) = h(g(f(x)))$ using the chain rule $\dfrac{dy}{dx} = \dfrac{dh}{dx}(g(f(x)))\cdot \dfrac{dg}{dx}(f(x))\cdot \dfrac{df}{dx}(x)$ $\dfrac{dy}{dx} = 2\sin(\tan(2x)) \cdot \cos(\tan(2x))\cdot 2 \sec^2(x) = 4 \sec^2(x) \cos(\tan(2x)) \sin(\tan(2x))$ Thanks from Seventy7 March 25th, 2017, 01:15 PM   #3
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Quote:
 Originally Posted by Seventy7 Find derivative of sin^2(tan2x) 2sin(tan2x)cos(tan2x)sec^2(2x)(2) =4sec^2(2x)sin(tan2x)cos(tan2x) First, are both the provided middle step and answer correct? If so... --------------------------------- I've been over this problem but can't get to that middle step. Is the first part using the combination of Power Rule and Chain Rule? I can't work it right. d/dx [sin^2(tan2x)] = 2sin(cos) du/dx [sin^2(tan2x)] how then to get to the middle step?
Mostly it's just switching the terms around.
$\displaystyle 2\color{brown}{\sin(\tan2x)}\cos(\tan2x)\sec^2(2x) (2)$

$\displaystyle 4\sec^2(2x)\color{brown} {\sin(\tan(2x))}\cos(\tan(2x))$

-Dan

Can someone show me how to format different colors and how to fix the LaTeX?

Last edited by skipjack; March 25th, 2017 at 03:07 PM. March 25th, 2017, 01:21 PM   #4
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Quote:
 Originally Posted by topsquark Mostly it's just switching the terms around. $\displaystyle 2\color{brown}{\sin(\tan2x)}\cos(\tan2x)\sec^2(2x) (2)$ $\displaystyle 4\sec^2(2x)\color{brown} {\sin(\tan(2x))}\cos(\tan(2x))$ -Dan Can someone show me how to format different colors?
$\color{red}{\text{red}}$
$\color{green}{\text{red}}$
$\color{cyan}{\text{red}}$
$\color{yellow}{\text{red}}$
$\color{black}{\text{red}}$
$\color{brown}{\text{red}}$
$\color{blue}{\text{red}}$

I don't know what the full vocab of colors is or whether there are additional modifiers you can use.

Last edited by skipjack; March 25th, 2017 at 03:08 PM. March 25th, 2017, 01:24 PM #5 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 $y = \sin^2(\tan\{x\}).$ $u = \tan(2x)\ and\ v = \sin(u) \implies y = v^2 \implies$ $\dfrac{dy}{dv} = 2v\ and\ \dfrac{dv}{du} = \cos(u)\ and\ \dfrac{du}{dx} = \sec^2(2x) * 2 = \dfrac{2}{\cos^2(2x)} \implies$ $\dfrac{dy}{dx} = \dfrac{dy}{dv} * \dfrac{dv}{du} * \dfrac{du}{dx} \implies$ $\dfrac{dy}{dx} = 2v * \cos(u) * \dfrac{2}{\cos^2(2x)} = \dfrac{2\{2\sin(u) * \cos(u)\}}{\cos^2(2x)} = \dfrac{2\sin(2u)}{\cos^2(2x)} = \dfrac{2\sin(2\tan\{2x\})}{\cos^2(2x)}.$ Thanks from Seventy7 Last edited by skipjack; March 25th, 2017 at 03:10 PM. March 25th, 2017, 01:42 PM #6 Member   Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 What I think I did wrong was trying to use the combination of chain rule and power rule. Another mistake was in the three functions: I had chosen f=sin^2(g); g=tan(u); and u=2x. I see now that the three layers of functions are f=g^2; g=sin(u); and u= tan2x Starting on outside and working inside. Derivative of x^2 is 2x; replace x with sin(tan2x) = 2sin(tan2x) Move to the middle function sin x. derivative is cos x, replace x with tan2x So far, I have 2sin(tan2x)cos(tan2x) Finally move to the third function: derivative of tan 2x; this is chain rule Derivative of tan 2x = sec^2 (2x) (2) Putting it althogether: 2sin(tan2x)cos(tan2x)sec^2(2x)(2) Then simplify. I think I've got it now. Thanks! Last edited by Seventy7; March 25th, 2017 at 01:52 PM. Reason: fixed three layers of functions March 25th, 2017, 02:04 PM   #7
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Quote:
 Originally Posted by Seventy7 What I think I did wrong was trying to use the combination of chain rule and power rule. Another mistake was in the three functions: I had chosen f=sin^2(g); g=tan(u); and u=2x. I see now that the three layers of functions are f=g^2; g=sin(u); and u= tan2x Starting on outside and working inside. Derivative of x^2 is 2x; replace x with sin(tan2x) = 2sin(tan2x) Move to the middle function sin x. derivative is cos x, replace x with tan2x So far, I have 2sin(tan2x)cos(tan2x) Finally move to the third function: derivative of tan 2x; this is chain rule Derivative of tan 2x = sec^2 (2x) (2) Putting it althogether: 2sin(tan2x)cos(tan2x)sec^2(2x)(2) Then simplify. I think I've got it now. Thanks!
Your original approach will work.

$g = \tan(u),\ u = 2x,\ and\ f = \sin^2(g) \implies$

$\dfrac{df}{dg} = 2\sin(g)\cos(g) = \sin(2g)\ and\ \dfrac{dg}{du} = \sec^2(u) = \dfrac{1}{\cos^2(u)}\ and\ \dfrac{du}{dx} = 2 \implies$

$\dfrac{df}{dx} = \dfrac{df}{dg} * \dfrac{dg}{du} * \dfrac{du}{dx} \implies$

$\dfrac{df}{dx} = \sin(2g) * \dfrac{1}{\cos^2(u)} * 2 = \dfrac{2\sin(2\tan\{2x\})}{\cos^2(2x)}.$

Last edited by skipjack; March 25th, 2017 at 03:11 PM. March 25th, 2017, 04:41 PM   #8
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Quote:
 Originally Posted by romsek $\color{red}{\text{red}}$ $\color{green}{\text{red}}$ $\color{cyan}{\text{red}}$ $\color{yellow}{\text{red}}$ $\color{black}{\text{red}}$ $\color{brown}{\text{red}}$ $\color{blue}{\text{red}}$ I don't know what the full vocab of colors is or whether there are additional modifiers you can use.
(Doh) I spent some time trying to find a font bar on the formatting menu. It never occurred to me to simply code it.

Thanks!

-Dan Tags derivative, figure, step or answer Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Angelwngs26 Elementary Math 8 April 11th, 2016 08:55 AM vrcat25 Number Theory 1 January 31st, 2016 11:33 AM SamSeymour Calculus 2 April 21st, 2014 02:07 PM coachcft Calculus 1 December 15th, 2012 10:05 AM unwisetome3 Calculus 5 October 17th, 2012 06:58 PM

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