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Not after an answer, just the method/procedure as I'm stumped...

We have the functionals:

$$ T[y] = \int_2^3 \left( 3\left| \frac{dy}{dx}\right|^2 - 8y \right)dx $$ $$ S[y] = \cosh(T[y]) $$

Now, to find the Gâteaux derivative of $S[y]$ my approach is to use the Chain Rule for $$ S[y] = \cosh \left( \int_2^3 \left( 3 \left|\frac{dy}{dx}\right|^2 - 8y \right) dx \right) $$

Would this be correct?

Or would I compute the Gâteaux Derivative of $T[y]$ first and then substitute into $S[y]$ and compute the Gâteaux Derivative of $S[y]$ next?

Any pointers would be appreciated, as I say I'm not really after the answer just the correct method.


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Yes it is enough to use the chain rule: If $X$ is the space of functions you are working (where I assume that you can derivate), then you have $T:X\rightarrow\mathbb{R}$ and $\cosh:\mathbb{R}\rightarrow\mathbb{R}$, so $S=\cosh\circ T:X\to\mathbb{R}$ and $\langle(\cosh\circ T)'(x),y\rangle=\cosh'(T(x))\langle T'(x),y\rangle=\sinh(T(x))\langle T'(x),y\rangle$.

Here $\langle\cdot,\cdot\rangle$ denotes duality.

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My suggestion is the following:

First, compute the Gateaux-derivative of $T$. Since $T$ is quadratic, this should be no problem.

Then, you could use a chain-rule for $S$ or you do it directly: $$ S[y + t \, dy] = \cosh( T[y + t \, dy ] ) = \cosh( T[y] + t \, T'[y] \, dy + o(t) ) $$ and then you can continue using the differentiability of $\cosh$.

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The chain rule for Gateaux derivatives takes a different form to the usual chain rule - see Bernhard 2005:

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