# Higher order derivatives using the chain rule

tompenny
Homework Statement:
I need to show that higher order derivatives is applicable on a rotation in the plane
Relevant Equations:
[u v] = [ cos θ -sin θ * [x y]
sin θ cos θ]

∂2f∂x2+∂2f∂y2=∂2fu2+∂2fv2
Mentor note: Fixed the LaTeX in the following
I have the following statement:

\begin{cases} u=x \cos \theta - y\sin \theta \\ v=x\sin \theta + y\cos \theta \end{cases}

I wan't to calculate:

$$\dfrac{\partial^2}{\partial x^2}$$

My solution for ##\dfrac{\partial^2}{\partial x^2}##

##\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}##

##\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} f = \left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right) f = \cos \theta \dfrac{\partial f}{\partial u} - \sin \theta\dfrac{\partial f}{\partial v}##.

Proceeding on to the second derivative I get:

##\dfrac{\partial^2}{\partial x^2} = \left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right)\left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right) = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} - 2\sin \theta \cos \theta \dfrac{\partial^2}{\partial u\partial v} + \sin^2 \theta \dfrac{\partial^2}{\partial v^2}##.

Similarly I get ##\dfrac{\partial^2}{\partial y^2} = \left( \sin \theta \dfrac{\partial}{\partial u} \cos \theta\dfrac{\partial}{\partial v} \right)\left( \sin\theta \dfrac{\partial}{\partial u} \cos \theta\dfrac{\partial}{\partial v} \right) = \sin^2 \theta \dfrac{\partial^2}{\partial u^2} - 2\sin \theta \cos \theta \dfrac{\partial^2}{\partial u\partial v} + \cos^2 \theta \dfrac{\partial^2}{\partial v^2}##.

But how do I calculate ##\dfrac{\partial^2}{\partial u^2}## , ##\dfrac{\partial^2}{\partial v^2}## ?

Any tips would be greatly appreciated:)

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tompenny
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2022 Award
Homework Statement:: I need to show that higher order derivatives is applicable on a rotation in the plane
Relevant Equations:: [u v] = [ cos θ -sin θ * [x y]
sin θ cos θ]

I have the following statement:

$$\begin{cases} u=x \cos \theta - y\sin \theta \\ v=x\sin \theta + y\cos \theta \end{cases}$$

My solution for $\dfrac{\partial^2}{\partial x^2}$

$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}$$

$$\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} f = \left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right) f = \cos \theta \dfrac{\partial f}{\partial u} - \sin \theta\dfrac{\partial f}{\partial v}$$

Proceeding on to the second derivative I get:

$$\dfrac{\partial^2}{\partial x^2} = \left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right)\left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right) = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} - 2\sin \theta \cos \theta \dfrac{\partial^2}{\partial u\partial v} + \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$.

Similarly I get $$\dfrac{\partial^2}{\partial y^2} = \left( \sin \theta \dfrac{\partial}{\partial u} \cos \theta\dfrac{\partial}{\partial v} \right)\left( \sin\theta \dfrac{\partial}{\partial u} \cos \theta\dfrac{\partial}{\partial v} \right) = \sin^2 \theta \dfrac{\partial^2}{\partial u^2} - 2\sin \theta \cos \theta \dfrac{\partial^2}{\partial u\partial v} + \cos^2 \theta \dfrac{\partial^2}{\partial v^2}$$.

But how do I calculate $$\dfrac{\partial^2}{\partial u^2}$$ , $$\dfrac{\partial^2}{\partial v^2}$$ ?

Any tips would be greatly appreciated:)

You don't have to calculate others. You can simplify what you have.

However, you have a mistake in the first line. This is wrong:

$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}$$

tompenny
tompenny
You don't have to calculate others. You can simplify what you have.

However, you have a mistake in the first line. This is wrong:

$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}$$

Thank you so much for helping me.. should the first line be:
$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} + \sin \theta\dfrac{\partial}{\partial v}$$ ??

Mentor
@tompenny, this site uses MathJax, which accepts some LaTeX stuff. The delimiters are pairs of $$characters (standalone LaTeX) or ## (inline). Science Advisor Homework Helper Gold Member 2022 Award Thank you so much for helping me.. should the first line be:$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} + \sin \theta\dfrac{\partial}{\partial v}$$?? Yes. What about$$\dfrac{\partial^2}{\partial x^2}$$. Hint: write everything out carefully. tompenny$$\dfrac{\partial^2}{\partial x^2} = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} + 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$Is that correct? :) Can you give me any tips on how to calculate$$\dfrac{\partial^2}{\partial u^2} $$MAny thanks for all your help:) PeroK Science Advisor Homework Helper Gold Member 2022 Award$$\dfrac{\partial^2}{\partial x^2} = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} + 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$Is that correct? :) Can you give me any tips on how to calculate$$\dfrac{\partial^2}{\partial u^2} $$MAny thanks for all your help:) Why not try$$\dfrac{\partial^2}{\partial y^2} $$tompenny then I get:$$\dfrac{\partial^2}{\partial y^2} = \sin^2 \theta \dfrac{\partial^2}{\partial u^2} - 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \cos^2 \theta \dfrac{\partial^2}{\partial v^2}$$Can I somewhat use this to calculate$$\dfrac{\partial^2}{\partial u^2} ?

Thank you:)