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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: 2,426 Thanks: 1314 
$y(x) = \sin^2(\tan(2x))$ 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 $\dfrac{dy}{dx} = \dfrac{dh}{dx}(g(f(x)))\cdot \dfrac{dg}{dx}(f(x))\cdot \dfrac{df}{dx}(x)$ $\dfrac{dy}{dx} = 2\sin(\tan(2x)) \cdot \cos(\tan(2x))\cdot 2 \sec^2(x) = 4 \sec^2(x) \cos(\tan(2x)) \sin(\tan(2x))$ 
March 25th, 2017, 01:15 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,151 Thanks: 875 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle 2\color{brown}{\sin(\tan2x)}\cos(\tan2x)\sec^2(2x) (2)$ $\displaystyle 4\sec^2(2x)\color{brown} {\sin(\tan(2x))}\cos(\tan(2x))$ 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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,426 Thanks: 1314  Quote:
$\color{green}{\text{red}}$ $\color{cyan}{\text{red}}$ $\color{yellow}{\text{red}}$ $\color{black}{\text{red}}$ $\color{brown}{\text{red}}$ $\color{blue}{\text{red}}$ 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 
$y = \sin^2(\tan\{x\}).$ $u = \tan(2x)\ and\ v = \sin(u) \implies y = v^2 \implies$ $\dfrac{dy}{dv} = 2v\ and\ \dfrac{dv}{du} = \cos(u)\ and\ \dfrac{du}{dx} = \sec^2(2x) * 2 = \dfrac{2}{\cos^2(2x)} \implies$ $\dfrac{dy}{dx} = \dfrac{dy}{dv} * \dfrac{dv}{du} * \dfrac{du}{dx} \implies$ $\dfrac{dy}{dx} = 2v * \cos(u) * \dfrac{2}{\cos^2(2x)} = \dfrac{2\{2\sin(u) * \cos(u)\}}{\cos^2(2x)} = \dfrac{2\sin(2u)}{\cos^2(2x)} = \dfrac{2\sin(2\tan\{2x\})}{\cos^2(2x)}.$ 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  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
$g = \tan(u),\ u = 2x,\ and\ f = \sin^2(g) \implies$ $\dfrac{df}{dg} = 2\sin(g)\cos(g) = \sin(2g)\ and\ \dfrac{dg}{du} = \sec^2(u) = \dfrac{1}{\cos^2(u)}\ and\ \dfrac{du}{dx} = 2 \implies$ $\dfrac{df}{dx} = \dfrac{df}{dg} * \dfrac{dg}{du} * \dfrac{du}{dx} \implies$ $\dfrac{df}{dx} = \sin(2g) * \dfrac{1}{\cos^2(u)} * 2 = \dfrac{2\sin(2\tan\{2x\})}{\cos^2(2x)}.$ Last edited by skipjack; March 25th, 2017 at 03:11 PM.  
March 25th, 2017, 04:41 PM  #8  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,151 Thanks: 875 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Thanks! Dan  

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