"Use differentials to approximate $\displaystyle \sin{ln(1.1)^{1.1}}$."

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- Oct 16th 2009, 04:17 PMxxlvhDifferentials?
"Use differentials to approximate $\displaystyle \sin{ln(1.1)^{1.1}}$."

- Oct 16th 2009, 04:25 PMadkinsjr
- Oct 16th 2009, 05:34 PMxxlvh
Not much at all. My prof skipped the topic altogether, and this question is due on monday. (He often teaches subjects after we have assignments on them)

I've scanned through the differentials section of my textbook but none of their examples looked like this.

I think I can find the actual differential, but doesn't there have to be an x and a dx value given? - Oct 16th 2009, 06:14 PMadkinsjr
Ok, finding differentials is easy:

$\displaystyle \frac{dy}{dx}=f'(x)$

So the differential $\displaystyle dy$ is just:

$\displaystyle dy=f'(x)dx$

This will help get you started:

Mathwords: Approximation by Differentials

In your case, you would be best to define your function ase $\displaystyle f(x)=sin(ln(x)^x)$, which can be written as $\displaystyle sin(xln(x))$

Use $\displaystyle \Delta x =0.1$ since you know that $\displaystyle (1)ln(1)=ln(1)=0$ is very close to $\displaystyle 1.1ln(1.1)$ - Oct 16th 2009, 06:15 PMadkinsjr
So if you know how to find the derivative of $\displaystyle y=sin(xln(x))$, then you should have no problem applying the formula: $\displaystyle dy=f'(x)dx$.

- Oct 16th 2009, 06:42 PMHallsofIvy
- Oct 16th 2009, 06:55 PMxxlvh
It is $\displaystyle \sin{ln(1.1)^{1.1}}$

- Oct 16th 2009, 07:27 PMadkinsjr
- Oct 16th 2009, 07:31 PMadkinsjr
I guess it would be better to write it as $\displaystyle (ln(1.1))^{1.1}$ if we aren't applying the exponent to the arguement itself, but to the entire value. So the approximation is around 0.0753 according to my calculator.

- Oct 16th 2009, 07:49 PMxxlvh
Would you mind checking my solution to make sure I didn't make any errors? It's my first time working with these.

$\displaystyle f(x) = \sin{ln(x)^x}$ with x = 1 and $\displaystyle \Delta x=0.1$

$\displaystyle dy = \cos{(xln(x)}(1+ln(x))dx$

$\displaystyle \sin{ln(1.1)^{1.1}} \approx f(x) + f'(x)\Delta x $

$\displaystyle = \sin{ln(1)^1} + \cos{(1*ln1)}(1+ln1)(0.1) $

$\displaystyle = 0.1$ - Oct 16th 2009, 08:14 PMadkinsjr
I'm still not very sure of the notation here. Does $\displaystyle ln(x)^x$ mean that we are taking the arguement of the logaritm to the power of x, or the logaritm itself.

For example, if I want to square the sine of x, I write $\displaystyle sin^2(x)$, I wouldn't write it like $\displaystyle sin(x)^2$, because that would imply that I am just squaring the arguement. Is the function $\displaystyle (ln(x))^x$ ??? - Oct 16th 2009, 08:23 PMxxlvh
I am not sure, I typed it exactly as it was given to me.