Higher order derivatives using the chain rule

In summary, the conversation discusses the calculation of higher order derivatives in the context of rotation in a plane. The given statement is transformed into a partial derivative equation, and the solution for the second derivative with respect to x is shown. The participant inquires about calculating the second derivatives with respect to u and v, to which the mentor provides a hint and suggests trying out the calculation for the second derivative with respect to y. The participant then successfully calculates the second derivative with respect to u using the pythagorean trigonometric identity.
  • #1
tompenny
15
3
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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  • #3
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  • #4
tompenny said:
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:

$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}$$
 
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  • #5
PeroK said:
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:
$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} + \sin \theta\dfrac{\partial}{\partial v}$$ ??
 
  • #6
@tompenny, this site uses MathJax, which accepts some LaTeX stuff. The delimiters are pairs of $$ characters (standalone LaTeX) or ## (inline).
 
  • #7
tompenny said:
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.
 
  • #8
$$\dfrac{\partial^2}{\partial x^2} = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} + 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$

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:)
 
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  • #9
tompenny said:
$$\dfrac{\partial^2}{\partial x^2} = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} + 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$

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 $$\dfrac{\partial^2}{\partial y^2} $$
 
  • #10
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:)
 
  • #11
What about adding those two together?
 
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  • #12
Hahaha.. I finally got it! the mixed partial cancels and if I use the pythagorean trigonometric identity on what's left I get the answer!

I can't explain how much you've made my day! <3
 
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Related to Higher order derivatives using the chain rule

What is the chain rule?

The chain rule is a mathematical rule used to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

How is the chain rule used to find higher order derivatives?

The chain rule can be applied multiple times to find higher order derivatives. Each time, the derivative of the outer function is multiplied by the derivative of the inner function.

Can the chain rule be used with any type of function?

Yes, the chain rule can be used with any type of function, including polynomial, exponential, logarithmic, and trigonometric functions.

What is the notation for higher order derivatives using the chain rule?

The notation for higher order derivatives using the chain rule is similar to that of first order derivatives, where the prime symbol (') is used. For example, the second derivative of a function f(x) would be written as f''(x).

Are there any special cases when using the chain rule for higher order derivatives?

Yes, there are some special cases when using the chain rule for higher order derivatives, such as when the inner function is a constant or when the outer function is a composite function itself. In these cases, the chain rule may need to be applied multiple times or other derivative rules may need to be used.

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