# Math Help - Two quick questions on the 'differential' notation

1. ## Two quick questions on the 'differential' notation

(1) Why is it OK to change the integrand's step like this: $\int \frac{dx}{x\log x}=\int\frac{d(\log x)}{\log x}=\log\log x + C$
What is the (geometrical or some other) intuition behind this?
This is from a (somewhat) rigorous analysis book ( $\epsilon$- $\delta$ proofs galore), and such integration-step-substitutions were being made without explanation.

(2) Let's say that $\alpha=f\left(x(t),y(t)\right)$. Please help me write $\frac{d\alpha}{dt}$ with differentials:
$\frac{d\alpha}{dt}=\frac{d}{dt}f(x(t),y(t))=f'(x(t ),y(t))\cdot\text{?}$
How do you continue?

2. ## Re: Two quick questions on the 'differential' notation

1) i don't think it is intuition.it is just integration by substitution.

if you say that some t=log(x)
then dt/dx=1/x

or dt=dx/x

d(\log x)=\frac{dx}{x}

now you can substitute dx/x with d(log(x))

3. ## Re: Two quick questions on the 'differential' notation

Thank you! Would you also know the answer to part (2)?

4. ## Re: Two quick questions on the 'differential' notation

sure!

2) $\frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))$

You get that from the chain rule. Chain rule - Wikipedia, the free encyclopedia

edit: i fixed the whole thing right up.

5. ## Re: Two quick questions on the 'differential' notation

Originally Posted by anonimnystefy
sure!

2) $\frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))$

You get that from the chain rule. Chain rule - Wikipedia, the free encyclopedia

my latex isn't working.
You have to use [ tex ] ... [ /tex ], not [ math] ... [ /math]

6. ## Re: Two quick questions on the 'differential' notation

Originally Posted by anonimnystefy
2) $\frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))$
Only now did I remember to quote and change the tags to [ TEX ] myself.

7. ## Re: Two quick questions on the 'differential' notation

Originally Posted by anonimnystefy
2) $\frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))$
How would one apply this to, say, the slope $\alpha$ of a parameterized curve $\alpha=\arctan\left(\frac{\dot{y}}{\dot{x}}\right)$ at the point $(x,y)$?

$\frac{d}{dt} \Big{(}\alpha\Big{)} = \frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \hdots$

Now, what is $\frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$? The final answer should be $d\alpha=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}dt$, so the $\frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$ apparently equals $\dot{x}\ddot{y}-\ddot{x}\dot{y}$. How do you get this?

8. ## Re: Two quick questions on the 'differential' notation

Originally Posted by courteous
How would one apply this to, say, the slope of a parameterized curve $\alpha=\arctan\left(\frac{\dot{y}}{\dot{x}}\right)$ at the point $(x,y)$?

$\frac{d}{dt} \Big{(}\alpha\Big{)} = \frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \hdots$

Now, what is $\frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$? The final answer should be $d\alpha=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}dt$, so the $\frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$ apparently equals $\dot{x}\ddot{y}-\ddot{x}\dot{y}$. How do you get this?
There's something strange I think, if you want to calculate:
$\frac{d}{dt}\left(\alpha\right)=\frac{d}{dt}\left[\arctan\left(\frac{y}{x}\right)\right]=0$
Because there's no variable $t$ in the function, that means $\alpha$ is a constant.

My other question, I'm not famimiar with the dots on the letters. What do they mean? ...

9. ## Re: Two quick questions on the 'differential' notation

$x$ and $y$ are both functions of the parameter $t$. Therefore, $\dot{x}=\frac{dx}{dt}$. Oh, and the dot is Newton's way of writing the derivative.

10. ## Re: Two quick questions on the 'differential' notation

Ok, Sorry, forget my answer, I was confused.

11. ## Re: Two quick questions on the 'differential' notation

Got slightly further: since $\frac{\dot{y}}{\dot{x}}=\frac{dy}{dx}$, we can rewrite $\frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big)= \frac{d}{dt}\Big(\frac{dy}{dx}\Big)=\frac{d}{dx} \Big(\frac{dy}{dx}\Big)\cdot\frac{dx}{dt}=y''(x) \cdot \dot{x}$

To put everything in one place:
$\frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \frac{1}{\dot{x}^2+\dot{y}^2} \cdot y''(x) \cdot \dot{x} = \hdots$

Is this going in the right direction? If so, then what is $y''(x)$ if you have $x(t)$ and $y(t)$?

PS. Also, when $x$ and $y$ are functions of $t$, does there always exist $y(x)$?

12. ## Re: Two quick questions on the 'differential' notation

Originally Posted by anonimnystefy
sure!

2) $\frac{d}{dt}f(x(t),y(t))=x'(t) \cdot y'(t) \cdot f'(x(t),y(t))$
I might mistake, but it doesn't seem right to me. What does $f'(a)$ mean ? If it's the differential of $f$ in $a$, then it's a linear application and your equality isn't homogeneous.

$\alpha' (t) = \mathrm d_{(x(t),y(t))}f(x'(t),y'(t))$ by a simple composition of $f$ and $t \mapsto (x(t),y(t))$.

If $f : \mathbb R^2 \rightarrow \mathbb R$, then we express it like that :

$\alpha' (t) = x'(t)\frac {\partial f}{\partial x}(x(t),y(t)) + y'(t)\frac{\partial f}{\partial y}(x(t),y(t))$

@courteous : can you precise your goal, because your function $\alpha = \arctan \left( \frac {\. x} {\.y} \right)$ is not of the form $t \mapsto f(x(t),y(t))$.

13. ## Re: Two quick questions on the 'differential' notation

Yes, what anonimnystefy wrote was wrong. If x and y are both functions of t and f is a function of x and y then
$\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$

14. ## Re: Two quick questions on the 'differential' notation

Originally Posted by pece
@courteous : can you precise your goal, because your function $\alpha = \arctan \left( \frac {\. x} {\.y} \right)$ is not of the form $t \mapsto f(x(t),y(t))$.
Hello pece! I want to differentiate the slope/angle $\alpha$ of a parameterized curve with regards to parameter $t$ ... that is, I'd like to find $\frac{d\alpha}{dt}=\frac{d}{dt}\arctan\Big(\frac{ \dot{y} }{ \dot{x} }\Big)$ at a point $(x(t),y(t))$.

EDIT: And so that everything is summed-up in this one post, this is what I've figured out so far (not necessarily correct though):
$\frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \frac{1}{\dot{x}^2+\dot{y}^2} \cdot y''(x) \cdot \dot{x} = \hdots$

Now, I don't know what to do with the $y''(x)=\frac{d^2y}{dx^2}$ term.

15. ## Re: Two quick questions on the 'differential' notation

I highly doubt that no one knows the answer ... HallsofIvy certainly knows it. Please, help ... what is $\frac{d}{dt}\Big(\alpha\Big)=\frac{d}{dt}\Big( \arctan(\frac{ \dot{y} }{ \dot{x} }) \Big)$? The complete question is in previous (#14) post.

PS. The final answer should be $\frac{d\alpha}{dt}=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}$.