y= e^sin 2x

Solution attempt:

lny = sin 2x

lny = 2sinxcosx

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- Mar 1st 2011, 06:13 PMFoxlionnatural differentiation
y= e^sin 2x

Solution attempt:

lny = sin 2x

lny = 2sinxcosx - Mar 1st 2011, 06:18 PMNOX Andrew
$\displaystyle \dfrac{d}{dx}\left[e^u\right] = e^u \times \dfrac{du}{dx}$

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

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

$\displaystyle y' = e^{\sin(2x)} \times 2\cos(2x) = 2e^{\sin(2x)}\cos(2x)$ - Mar 1st 2011, 06:26 PMintegral
$\displaystyle y= e^{sin 2x}$

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

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

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

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

$\displaystyle y=e^{sin 2x}$

$\displaystyle \frac{dy}{dx}=2e^{sin 2x}\cos(2x)$ - Mar 1st 2011, 07:00 PMFoxlion
Then, $\displaystyle \dfrac{du}{dx} = 2\cos(2x)$.

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 $\displaystyle \dfrac{du}{dx}$?

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

did I just make a mistake? - Mar 1st 2011, 07:09 PMintegral
Product rule applies when you have the form: $\displaystyle f(x)=p(x)q(x)$. (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: $\displaystyle f(x)=p(q(x))$

where p(x)=sin(x) and q(x)=2x - Mar 1st 2011, 07:12 PMFoxlion
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.