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#### DeusAbscondus

##### Active member

- Jun 30, 2012

- 176

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

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

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