$F(x,y)=sin(||f(x)+y||)^2$ - find the $DF (x, y) (u, v)$ through $Df (x) (u).$

Question

Let $f$: $\mathbb{R^n}$ → $\mathbb{R^n}$ differentiable mapping and $F$: $\mathbb{R^n x R^n} \to \mathbb{R}$ mapping the default formula: $$F(x,y)=sin||f(x)+y||^2$$

Find the $DF (x, y) (u, v)$ through $Df (x) (u).$
My work:

Idea is to write $F$ like compostion of many function that we know how their deerivates look.
Let's $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$(this are linnear operators)
$g1=f\circ\pi_1+\pi_2$
$F=sin||g1||^2$
$D_{g1}(x,y)(u,v)=Df(\pi_{1}(x,y)[Df(\pi_1(x,y)(u,v))] + D\pi_2(x,y)(u,v) =D_f(x)*u+v$
Is this ok step?
$D_F(x,y)(u,v)=cos(||f(x)+y||^2)*2$

Answer 1

We can write $F(x,y) = g\circ h \circ w (x,y),$ where $g(t) = \sin t$ for $t \in \mathbb R,$ $h(x) = |x|^2 $ for $x\in \mathbb R^n,$ $w(x,y) = f(x) + y$ for $(x,y)\in \mathbb R^n\times \mathbb R^n.$ By the chain rule,

$$\tag 1 DF(x,y) = Dg(h \circ w (x,y))\circ Dh(w(x,y))\circ Dw(x,y).$$

Now

$$\tag 2 Dg(t)(v) = \cos t\cdot v, v\in \mathbb R,\,\,\text { and }\,\, Dh(x)(v) = 2\langle x,v\rangle, v\in \mathbb R^n.$$

To calculate $Dw(x,y),$ let's do what you did, namely write $w = f\circ \pi_1 + \pi_2,$ where $\pi_1(x,y) = x,\pi_2(x,y)= y.$ Because $\pi_1, \pi_2$ are linear, we have $D\pi_1 = \pi_1, D\pi_2 = \pi_2$ everywhere. Thus, again using the chain rule,

$$\tag 3 Dw(x,y) = D(f\circ \pi_1 + \pi_2)(x,y) = Df(x)\circ \pi_1 + \pi_2.$$

We can now calculate $DF(x,y)(u,v).$ First,

$$Dw(x,y)(u,v) = Df(x)(u) +v.$$

Next,

$$Dh(w(x,y))(Df(x)(u) +v) = 2\langle (w(x,y),Df(x)(u) +v\rangle = 2\langle f(x) + y,Df(x)(u) +v\rangle.$$

Now we just multiply the last extression by $\cos (h\circ w(x,y)) = \cos (|f(x)+y|^2)$ to get the final answer

$$\cos (|f(x)+y|^2)\cdot 2\langle f(x) + y,Df(x)(u) +v\rangle.$$

Answer 2

You have one mistake. Let me start from the beginning. We are composing the following maps in order

$$f\times Id:(x,y)\to(f(x),y)$$

$$(\cdot + \cdot):(u,v)\to u+v$$

$$||\cdot||^2:t\to||t||^2$$

$$\sin:r\to\sin(r)$$

on the following spaces

$$\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^n\to\mathbb{R}\to\mathbb{R}$$

sending

$$(x,y)\to(f(x),y)\to f(x)+y\to ||f(x)+y||^2\to\sin(||f(x)+y||^2)$$

The differentials are

$$Df\times Id = \begin{bmatrix} Df|_x & 0 \\ 0 & I_n \end{bmatrix}$$

$$(I_n\ I_n)$$

$$2(t_1,\dots,t_n)$$

$$\cos(r)$$

Keep in mind the latter two differentials must be evaluated at $t=f(x)+y$ and $r=||f(x)+y||^2$ respectively.

Putting it together

$$DF|_{(x,y)}(u,v)=\cos(||f(x)+y||^2)\cdot 2 (f(x)+y))\cdot(I_n\ I_n)\cdot\begin{bmatrix} Df|x & 0 \\ 0 & I_n \end{bmatrix}\binom{u}{v}$$

Note that $(f(x)+y)$ is supposed to be a row matrix here.

$$=2\cos(||f(x)+y||^2)\ \cdot$$