My Math Forum Derivative with step/answer; can't figure why?

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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!

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

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

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

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

Last edited by skipjack; March 25th, 2017 at 03:11 PM.

March 25th, 2017, 04:41 PM   #8
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Math Focus: Wibbly wobbly timey-wimey stuff.
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

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