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March 25th, 2017, 12:41 PM   #1
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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.
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March 25th, 2017, 01:10 PM   #2
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$y(x) = \sin^2(\tan(2x))$

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
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March 25th, 2017, 01:15 PM   #3
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Quote:
Originally Posted by Seventy7 View Post
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?
Thanks from Seventy7

Last edited by skipjack; March 25th, 2017 at 03:07 PM.
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March 25th, 2017, 01:21 PM   #4
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Quote:
Originally Posted by topsquark View Post
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.
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March 25th, 2017, 01:24 PM   #5
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$y = \sin^2(\tan\{x\}).$

$u = \tan(2x)\ and\ v = \sin(u) \implies y = v^2 \implies$

$\dfrac{dy}{dv} = 2v\ and\ \dfrac{dv}{du} = \cos(u)\ and\ \dfrac{du}{dx} = \sec^2(2x) * 2 = \dfrac{2}{\cos^2(2x)} \implies$

$\dfrac{dy}{dx} = \dfrac{dy}{dv} * \dfrac{dv}{du} * \dfrac{du}{dx} \implies$

$\dfrac{dy}{dx} = 2v * \cos(u) * \dfrac{2}{\cos^2(2x)} = \dfrac{2\{2\sin(u) * \cos(u)\}}{\cos^2(2x)} = \dfrac{2\sin(2u)}{\cos^2(2x)} = \dfrac{2\sin(2\tan\{2x\})}{\cos^2(2x)}.$
Thanks from Seventy7

Last edited by skipjack; March 25th, 2017 at 03:10 PM.
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March 25th, 2017, 01:42 PM   #6
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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
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March 25th, 2017, 02:04 PM   #7
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Quote:
Originally Posted by Seventy7 View Post
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.

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

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

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

$\dfrac{df}{dx} = \sin(2g) * \dfrac{1}{\cos^2(u)} * 2 = \dfrac{2\sin(2\tan\{2x\})}{\cos^2(2x)}.$
Thanks from Seventy7

Last edited by skipjack; March 25th, 2017 at 03:11 PM.
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March 25th, 2017, 04:41 PM   #8
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Quote:
Originally Posted by romsek View Post
$\color{red}{\text{red}}$
$\color{green}{\text{red}}$
$\color{cyan}{\text{red}}$
$\color{yellow}{\text{red}}$
$\color{black}{\text{red}}$
$\color{brown}{\text{red}}$
$\color{blue}{\text{red}}$

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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