1. ## Differentials?

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

2. Originally Posted by xxlvh
"Use differentials to approximate $\sin{ln(1.1)^{1.1}}$."
How far can you take this? Do you know how to find differentials in general?

3. 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?

4. 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:

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

So the differential $dy$ is just:

$dy=f'(x)dx$

This will help get you started:

Mathwords: Approximation by Differentials

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

Use $\Delta x =0.1$ since you know that $(1)ln(1)=ln(1)=0$ is very close to $1.1ln(1.1)$

5. So if you know how to find the derivative of $y=sin(xln(x))$, then you should have no problem applying the formula: $dy=f'(x)dx$.

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

7. It is $\sin{ln(1.1)^{1.1}}$

8. Originally Posted by HallsofIvy
Is the formula $\sin{ln(1.1)^{1.1}}$ or is it $\sin{ln(1.1^{1.1})}$? What you wrote was the first but what adkinsjr gave is equivalent to the second.
lol, I'm confused now. Is $ln(1.1)^{1.1}$ The same as taking $ln(1.1)$ and then raising that value to the power of 1.1?

On the other hand we might have $ln(1.1^{1.1})=1.1ln(1.1)$ right?

9. I guess it would be better to write it as $(ln(1.1))^{1.1}$ 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.

10. Would you mind checking my solution to make sure I didn't make any errors? It's my first time working with these.

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

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

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

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

$= 0.1$

11. 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.

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

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

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

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

$= 0.1$
I'm still not very sure of the notation here. Does $ln(x)^x$ 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 $sin^2(x)$, I wouldn't write it like $sin(x)^2$, because that would imply that I am just squaring the arguement. Is the function $(ln(x))^x$ ???

12. I am not sure, I typed it exactly as it was given to me.