Question

quick question on differential calculus. ? in thread.

Answer 1

bare with me while i quickly type it.

Answer 2

given that g(r,theta)=f(rcos(theta), rsin(theta), if i were computing the derivatives, could i do dg/dtheta=df/dtheta, where d is partial deriv.

Answer 3

im trying to figure out how to write d^2g/dtheta^2 in terms of partial derivatives of f, the way i am thinking i can do it is using chain rule?

Answer 4

You should think of it as \(f(x(r,\theta),y(r,\theta))\)

Answer 5

What is the partial derivative you want to find?

Answer 6

see my post above does it make sense and am i on the right lines?

Answer 7

bare with me i will type it clearer.

Answer 8

\[\frac{dg^2}{d\theta^2}\] in terms of partial derivatives of f. And of course the d should be the part. deriv. but im not the sure the latex for it is

Answer 9

\[
\frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} +\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}
\]

Answer 10

because i was thinking, not sure im right, is say u=rcos(theta), and w=rsin(theta), the using chain rule dg/dtheta= dg/dumeult. du/dtheta + dg/dw mult. dw/dtheta right?

Answer 11

\[
\frac{\partial g}{\partial \theta} = \frac{\partial f}{\partial \theta}
\]

Answer 12

when it comes to taking the second derivative, how would i do it?

Answer 13

\[
\frac{\partial^2 f}{\partial \theta^2} =\frac{d}{d\theta}\left[ \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} +\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}
\right] \]Then use the product rule I suppose.

Answer 14

\[
\frac{d}{d\theta }\frac{\partial f}{\partial x} = \frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial \theta}
\]

Answer 15

what is the latex for partial derivative?

Answer 16

\frac{\partial F}{\partial X}

Answer 17

You can look at other people's latex too.

Answer 18

can you? how?

Answer 19

Right click > Show Math As > Tex Commands

Answer 20

\[
\frac{d}{d\theta }\left[\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}\right] = \frac{\partial^2 f}{\partial x^2}\left(\frac{\partial x}{\partial \theta}\right)^2+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial \theta^2}
\]

Answer 21

Quite a lot....\[
\frac{\partial^2 f}{\partial \theta^2} = \frac{\partial^2 f}{\partial x^2}\left(\frac{\partial x}{\partial \theta}\right)^2+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial \theta^2}+\frac{\partial^2 f}{\partial y^2}\left(\frac{\partial y}{\partial \theta}\right)^2+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial \theta^2}
\]

Answer 22

There is a difference between squaring the first derivative and the second derivative.

Answer 23

oh ok. so \[\frac{d}{d\theta }[\frac{\partial f}{\partial x} \cdot -r\sin\theta ] = \frac{\partial^2 f}{\partial x^2}\cdot -r\sin\theta+\frac{\partial f}{\partial x}\cdot -r\cos\theta\]

Answer 24

Yeah, that's the the product rule/chain rule.

Answer 25

But it should be squared...

Answer 26

oh ok? how comes it should be squared?

Answer 27

\[
\frac{d}{d\theta }\left[\frac{\partial f}{\partial x} \cdot -r\sin\theta \right] = \left(\underbrace{\frac{\partial^2 f}{\partial x^2}\cdot -r\sin\theta}_{\text{chain rule}}\right)\cdot(-r\sin\theta)+\frac{\partial f}{\partial x}\cdot -r\cos\theta
\]

Answer 28

Wait... hmmm

Answer 29

No, that looks about right...

Answer 30

\[\begin{split}
\frac{d}{d\theta }\left[\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}\right] &=\frac{d}{d\theta}\left[\frac{\partial f}{\partial\theta}\right]\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial x}\frac{d}{d\theta}\left[\frac{\partial x}{\partial \theta}\right] \\
&= \left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial \theta}\right)\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial \theta^2} \\
&=\frac{\partial^2 f}{\partial x^2}\left(\frac{\partial x}{\partial \theta}\right)^2+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial \theta^2}
\end{split}\]

Answer 31

@TedG get it now?

Answer 32

yh Thank you for that, helped alot.

Answer 33

@wio one last quick question see the beginning of the RHS, how is it \[\frac{d}{d\theta}\left[\frac{\partial f}{\partial\theta}\right]\] and not \[\frac{d}{d\theta}\left[\frac{\partial f}{\partial x}\right]\] from the LHS