Differentiation under the integral sign for a $L^1$-valued function (shape derivative)

Let

• $dinmathbb N$;
• $Usubseteqmathbb R^d$ be open and $$mathcal A:={Omegasubseteq U:Omegatext{ is bounded and open and }partialOmegatext{ is of class }C^{0,:1}};$$
• $E:=bigcup_{Omegainmathcal A}L^1(Omega)$;
• $y:mathcal Ato E$ with $$y(Omega)in L^1(Omega);;;text{for all }Omegainmathcal A;tag1$$
• $tau>0$ and $T_t:Uto U$ be a $C^1$-diffeomorphism for $tin[0,infty)$ with $T_0=operatorname{id}_D$ and $det{rm D}T_t(x)>0$ for all $(t,x)in[0,tau)times U$;
• $V:=bigcup_{tin[0,:tau)}T_t(U)$
• $v:[0,tau)times Vtomathbb R^d$ be differentiable in the second argument with $$vleft(t,T_t(x)right)=frac{partial T}{partial t}(t,x);;;text{for all }(t,x)in[0,tau)times U;tag2$$
• $Omegainmathcal A$ and $Omega_t:=T_t(Omega)$ for $tin[0,tau)$;
• $lambda$ denote the Lebesgue measure on $mathcal B(mathbb R)$;
• $sigma_{partialOmega}$ denote the surface measure on $mathcal B(partialOmega)$;
• $nu_{partialOmega}$ denote the unit outer normal field on $partialOmega$.

There is the following well-known result: If $fin L^1(V)$ is differentiable, then the functional $$mathcal F(A):=int_Af:{rm d}lambda^{otimes d};;;text{for }Ainmathcal A$$ is "shape differentiable at $Omega$ in direction $v$", i.e. $$frac{mathcal F(Omega_t)-mathcal F(Omega_0)}txrightarrow{tto0+}int_{partialOmega}flangle v_0,nu_{partialOmega}rangle:{rm d}sigma_{partialOmega}tag3.$$

Now let $Y:[0,tau)to E_d$ with $$left.Y(t)right|_{Omega_t}=y(Omega_t);;;text{for all }tin[0,tau).$$ Then $y$ is called shape differentiable at $Omega$ in direction $v$ if $Y$ is Fréchet differentiable at $0$. In that case, $$y'(Omega)(v):=left.Y'(0)right|_{Omega}tag4.$$

How can we show that $(4)$ does not depend on the choice of $Y$?

Let $Y_i$ be two choices of $Y$. Clearly, $$0=int_{Omega_t}(Y_1(t)-Y_2(t))varphi:{rm d}lambda^{otimes d}tag5$$ for all $varphiin C_c^infty(U)$.

But why does differentiating $(5)$ with respect to $t$ yield $$0=int_Omega(Y_1′(0)-Y_2′(0))varphi:{rm d}lambda^{otimes d}+int_{partialOmega}(Y_1(0)-Y_2(0))varphilangle v_0,nu_{partialOmega}rangle:{rm d}sigma_{partialOmega}tag5?$$

The second term on the right-hand side of $(5)$ seems to be an application of $(3)$ for $f=Y_1(0)-Y_2(0))varphi$. But where does the first term come from? It seems to be some kind of product rule which is applied here …