natural differentiation

**Foxlion** · March 1st 2011, 06:13 PM

y= e^sin 2x

Solution attempt:

lny = sin 2x

lny = 2sinxcosx

**NOX Andrew** · March 1st 2011, 06:18 PM

$\dfrac{d}{dx}\left[e^u\right] = e^u \times \dfrac{du}{dx}$

For $y = e^{\sin(2x)}$ , let $u = \sin(2x)$ . Then, $\dfrac{du}{dx} = 2\cos(2x)$ .

Now, it is only a matter of substituting the values of $u$ and $\dfrac{du}{dx}$ into the first equation at the top.

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

**integral** · March 1st 2011, 06:26 PM

$y= e^{sin 2x}$

$\ln(y)=\sin(2x)$

$\frac{dy}{ydx}=\frac{d}{dx}\left [ \sin(2x) \right ]$ implicite differentiation.

$\frac{dy}{ydx}=2\cos(2x)$ chain rule

$\frac{dy}{dx}=2y\cos(2x)$

$y=e^{sin 2x}$

$\frac{dy}{dx}=2e^{sin 2x}\cos(2x)$

**Foxlion** · March 1st 2011, 07:00 PM

Then, $\dfrac{du}{dx} = 2\cos(2x)$ .

Both of you have the correct answer, I get the use of the chain rule and it looks very clean but could you also use the product rule for $\dfrac{du}{dx}$ ?

Thus making $\dfrac{du}{dx}=2\sin+2x\cos$

did I just make a mistake?

**integral** · March 1st 2011, 07:09 PM

Product rule applies when you have the form: $f(x)=p(x)q(x)$ . (which you don't have here)

I am not sure where you are geting 2sin(2x)+2cos(2x) but if you will notice, the form of f(x) needing to use the chain rule is: $f(x)=p(q(x))$

where p(x)=sin(x) and q(x)=2x

**Foxlion** · March 1st 2011, 07:12 PM

Thank you integral, was confused as to which was the inside/ outside function. In fact I am rather ignorant of the transcendental functions altogether as I looked at sin2x as the product of sin and 2x.

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