"Use differentials to approximate $\displaystyle \sin{ln(1.1)^{1.1}}$."
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?
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 $\displaystyle 1.1ln(1.1)$
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) $
$\displaystyle = 0.1$
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 $\displaystyle (ln(x))^x$ ???