MHB Trig identity problem embedded in chain-rule myopia

[DeusAbscondus]

I'm having trouble understanding a trig identity and only include it here (rather than in trig forum) as it touches on a -broader- derivative problem.

Here it is:
$$\frac{d}{dx} \ e^{sin^2(x)}=e^{sin^2(x)}\cdot 2sin(x)cos(x)$$
$$=e^{sin^2(x)}\cdot \ sin(2x)$$

I have attached a proof of the derivation of sin^2(x) but I don't understand it, quite frankly. Especially "where u=sin(x)" I guess, because I've worked hard at internalizing the simple "truth"(?) that the brackets contain the "argument" which is being substituted by "u", but here, it is u=sin(x), then, following from that, $\frac{d}{dx} u^2=2u$, which i can accept if I first swallow $u=sin(x)$, but I can't, because I can't get over the hurdle of seeing $()$ as the cue for substituting with "u"

Edit: 22:55 GMT+10
** Does this get me any closer $$2u \cdot u'$$

Can anyone see my problem and lead me out of this quagmire so that I can proceed with the grunt work I need to do on the trig identities (remedial work, I know, but this is how I seem to be proceeding)

Sigh,
DeusAbs

 

[SuperSonic4]

I'm having trouble understanding a trig identity and only include it here (rather than in trig forum) as it touches on a -broader- derivative problem.

Here it is:
$$\frac{d}{dx} \ e^{sin^2(x)}=e^{sin^2(x)}\cdot 2sin(x)cos(x)$$
$$=e^{sin^2(x)}\cdot \ sin(2x)$$

I have attached a proof of the derivation of sin^2(x) but I don't understand it, quite frankly. Especially "where u=sin(x)" I guess, because I've worked hard at internalizing the simple "truth"(?) that the brackets contain the "argument" which is being substituted by "u", but here, it is u=sin(x), then, following from that, $\frac{d}{dx} u^2=2u$*[/color], which i can accept if I first swallow $u=sin(x)$, but I can't, because I can't get over the hurdle of seeing $()$ as the cue for substituting with "u"

Edit: 22:55 GMT+10
** Does this get me any closer $$2u \cdot u'$$

Can anyone see my problem and lead me out of this quagmire so that I can proceed with the grunt work I need to do on the trig identities (remedial work, I know, but this is how I seem to be proceeding)

Sigh,
DeusAbs

edit: If you're using a pseudonym you might want to black out the name in the top right corner of the screenshot.

* Nitpicking I'm sure but you're finding $\frac{d}{du}$ instead of $\frac{d}{dx}$You can use any substitution you like really although you do need to take care when finding $\frac{du}{dx}$. To make this easier to see let $y = e^{\sin^2(x)}$ (which also brings in the possibility of logarithmic differentiation). In your case having du inside the sin function doesn't really help since you can't go anywhere with it

$y = e^{\sin^2(x)}$

If we let $ u = \sin^2(x)$ then we end up with $\dfrac{du}{dx} = 2\sin(x)\cos(x) \text{ and } y = e^u$

edit - removed an "a" from this expression

By the chain rule $\frac{dy}{dx} = \dfrac{dy}{du} \cdot \frac{du}{dx}$

$\dfrac{dy}{dx} = e^u \cdot 2\sin(x)\cos(x) = e^{\sin^2(x)} \cdot 2\sin(x)\cos(x)$

=========================

It may be easier for you to solve this particular problem using logarithmic differentiation:
Let $y = e^{\sin^2(x)}\ \rightarrow \ \ln(y) = \sin^2(x) \ \rightarrow \ \frac{y'}{y} = 2\sin(x)\cos(x) \ \rightarrow \ y' = e^{\sin^2(x)} \cdot 2\cos(x)\sin(x)$

============================

Another way to look at it is to use the double angle identity for cos.
$\cos(2x) = 1-2\sin^2(x) \rightarrow \sin^2(x) = \frac{1}{2} - \frac{\cos(2x)}{2}$

$\dfrac{d}{dx}e^{\sin^2(x)} = \dfrac{d}{dx}e^{1/2 - \cos(2x)/2} $

Using the chain rule should hopefully be easier here without a squared term

$\dfrac{d}{dx}e^{1/2 - \cos(2x)/2} = e^{1/2 - \cos(2x)/2} \cdot \dfrac{d}{dx}- \cos(2x)/2 = e^{1/2 - \cos(2x)/2} \cdot \sin(2x)/2 \cdot 2 = e^{1/2 - \cos(2x)/2} \cdot \sin(2x)$

The value of the exponent is of course equal to sin^2(x) so feel free to back-sub
$e^{\sin^2(x)}\sin(2x)$

 

[DeusAbscondus]

edit: If you're using a pseudonym you might want to black out the name in the top right corner of the screenshot.

* Nitpicking I'm sure but you're finding $\frac{d}{du}$ instead of $\frac{d}{dx}$You can use any substitution you like really although you do need to take care when finding $\frac{du}{dx}$. To make this easier to see let $y = e^{\sin^2(x)}$ (which also brings in the possibility of logarithmic differentiation). In your case having du inside the sin function doesn't really help since you can't go anywhere with it

$y = e^{\sin^2(x)}$

If we let $ u = asin^2(x)$ then we end up with $\dfrac{du}{dx} = 2\sin(x)\cos(x) \text{ and } y = e^u$

By the chain rule $\frac{dy}{dx} = \dfrac{dy}{du} \cdot \frac{du}{dx}$

$\dfrac{dy}{dx} = e^u \cdot 2\sin(x)\cos(x) = e^{\sin^2(x)} \cdot 2\sin(x)\cos(x)$

=========================

It may be easier for you to solve this particular problem using logarithmic differentiation:
Let $y = e^{\sin^2(x)}\ \rightarrow \ \ln(y) = \sin^2(x) \ \rightarrow \ \frac{y'}{y} = 2\sin(x)\cos(x) \ \rightarrow \ y' = e^{\sin^2(x)} \cdot 2\cos(x)\sin(x)$

============================

Another way to look at it is to use the double angle identity for cos.
$\cos(2x) = 1-2\sin^2(x) \rightarrow \sin^2(x) = \frac{1}{2} - \frac{\cos(2x)}{2}$

$\dfrac{d}{dx}e^{\sin^2(x)} = \dfrac{d}{dx}e^{1/2 - \cos(2x)/2} $

Using the chain rule should hopefully be easier here without a squared term

$\dfrac{d}{dx}e^{1/2 - \cos(2x)/2} = e^{1/2 - \cos(2x)/2} \cdot \dfrac{d}{dx}- \cos(2x)/2 = e^{1/2 - \cos(2x)/2} \cdot \sin(2x)/2 \cdot 2 = e^{1/2 - \cos(2x)/2} \cdot \sin(2x)$

The value of the exponent is of course equal to sin^2(x) so feel free to back-sub
$e^{\sin^2(x)}\sin(2x)$

Simply Super!, Super.
There is substance herein to keep me chewing down hard for days, on real meat for real budding mathematicians! (Nerd)