Calculating $\text{D}g$ of $g(x,y) = \int_\frac1x^1\frac1t\exp(t^3x^2y)\text{d}t$ Let $g:(1,\infty)^2\to\mathbb{R}$ be given by 

$$g(x,y) = \int_\frac1x^1\frac1t\exp(t^3x^2y)\text{d}t.$$

How can I calculate $\text{D}g$ using parameter-dependent integrals?
 A: Since $g$ is defined on the open subset $U = ]1,+\infty[^{2}$ of $\mathbb{R}^{2}$, its differential is the map $Dg \, : \, a=(x,y) \, \longmapsto \, D_{a}g$ where $D_{a}g$ is a linear map from $\mathbb{R}^{2}$ to $\mathbb{R}$ defined by :
$$ \forall h \in \mathbb{R}^{2}, \, h=\begin{bmatrix} h_{1} \\ h_{2} \end{bmatrix}, \,  \, D_{a}g(h) = h_{1} \frac{\partial g}{\partial x}(a) + h_{2} \frac{\partial g}{\partial y}(a). $$
So, we have to compute the partial derivatives $\displaystyle \frac{\partial g}{\partial x}$ and $\displaystyle \frac{\partial g}{\partial y}$. 
To do so, let $G \, : \, ]0,1[ \times ]1,+\infty[^{2} \, \longrightarrow \, \mathbb{R}$ such that :
$$ \forall (u,x,y) \in \big( ]0,1[ \times ]1,+\infty[^{2} \big), \, G(u,x,y) = \int_{u}^{1} \frac{1}{t} \exp(t^{3}x^{2}y) \; dt $$
Note that :
$$ \forall (x,y) \in ]1,+\infty[^{2}, \, g(x,y) = G\bigg( \frac{1}{x},x,y \bigg) \tag{$\star$} $$
Applying the chain rule to $(\star)$ gives :
$$ \frac{\partial g}{\partial x}(x,y) = -\frac{1}{x^{2}} \frac{\partial G}{\partial u}\bigg( \frac{1}{x},x,y \bigg) + \frac{\partial G}{\partial x}\bigg( \frac{1}{x},x,y \bigg) \tag{1}$$
and
$$ \frac{\partial g}{\partial y}(x,y) = \frac{\partial G}{\partial y}\bigg( \frac{1}{x},x,y \bigg) \tag{2} $$
Now, we need to compute the partial derivatives $\displaystyle \frac{\partial G}{\partial u}$, $\displaystyle \frac{\partial G}{\partial x}$ and $\displaystyle \frac{\partial G}{\partial y}$.
Using the fundamental theorem of calculus, we can differentiate the identity $\displaystyle G(u,x,y) = - \int_{1}^{u} \frac{1}{t} \exp(t^{3}x^{2}y) \; dt$ and we get :
$$ \frac{\partial G}{\partial u}(u,x,y) = - \frac{1}{u} \exp( u^{3}x^{2}y) $$
Using the theorem of differentiation under the integral sign, we get :
$$ \frac{\partial G}{\partial x}(u,x,y) = - 2xy \int_{1}^{u} t^{2} \exp(t^{3}x^{2}y) \; dt $$
and 
$$ \frac{\partial G}{\partial y}(u,x,y) = - x^{2} \int_{1}^{u} t^{2} \exp(t^{3}x^{2}y) \; dt $$
As a consequence :
$$ \boxed{\displaystyle \frac{\partial g}{\partial x}(x,y) = \frac{1}{x} \exp\bigg( \frac{y}{x} \bigg) + 2xy \int_{\frac{1}{x}}^{1} t^{2} \exp(t^{3}x^{2}y) \; dt} $$
and 
$$ \boxed{\displaystyle \frac{\partial g}{\partial y}(x,y) = x^{2} \int_{\frac{1}{x}}^{1} t^{2} \exp(t^{3}x^{2}y) \; dt} $$
