# Differentials?

• Oct 16th 2009, 04:17 PM
xxlvh
Differentials?
"Use differentials to approximate $\displaystyle \sin{ln(1.1)^{1.1}}$."
• Oct 16th 2009, 04:25 PM
Quote:

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

How far can you take this? Do you know how to find differentials in general?
• Oct 16th 2009, 05:34 PM
xxlvh
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 PM
Quote:

Originally Posted by xxlvh
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 $\displaystyle 1.1ln(1.1)$
• Oct 16th 2009, 06:15 PM
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 PM
HallsofIvy
Quote:

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

Is the formula $\displaystyle \sin{ln(1.1)^{1.1}}$ or is it $\displaystyle \sin{ln(1.1^{1.1})}$? What you wrote was the first but what adkinsjr gave is equivalent to the second.
• Oct 16th 2009, 06:55 PM
xxlvh
It is $\displaystyle \sin{ln(1.1)^{1.1}}$
• Oct 16th 2009, 07:27 PM
Quote:

Originally Posted by HallsofIvy
Is the formula $\displaystyle \sin{ln(1.1)^{1.1}}$ or is it $\displaystyle \sin{ln(1.1^{1.1})}$? What you wrote was the first but what adkinsjr gave is equivalent to the second.

lol, I'm confused now. Is $\displaystyle ln(1.1)^{1.1}$ The same as taking $\displaystyle ln(1.1)$ and then raising that value to the power of 1.1?

On the other hand we might have $\displaystyle ln(1.1^{1.1})=1.1ln(1.1)$ right?
• Oct 16th 2009, 07:31 PM
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 PM
xxlvh
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 PM
Quote:

Originally Posted by xxlvh
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 $\displaystyle (ln(x))^x$ ???
• Oct 16th 2009, 08:23 PM
xxlvh
I am not sure, I typed it exactly as it was given to me.