Calculating the differential function of $f(x,y) = \int^{x+y}_{0}\phi (t)\,dt$?

Question

Let $\phi$ be a continuous function from $\mathbb{R}$ to $\mathbb{R}$ and $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $$f(x,y) = \int^{x+y}_{0}\phi(t)\,dt.$$ Calculate its differential function at any point $(x,y) \in \mathbb{R}^2$ along any vector $(h,k) \in \mathbb{R}^2$.

So, in multivariable calculus, the different notations in different languages can be quite confusing. So to make it clear, what I am asked to calculate is $df_{(x,y)}(h,k)$ which by definition is $$ df_{(x,y)}(h,k) = \underset{t \rightarrow 0}{\lim}\frac{f((x,y) + (th,tk)) - f(x,y)}{t}.$$

Now,for a fixed $t \in \mathbb{R}^*$, I find that:

$$\frac{f((x,y) + (th,tk)) - f(x,y)}{t} =\frac{1}{t}\left( \int^{x+y}_{0}\phi(z)\,dz + \int^{x+y + th + tk}_{x+y}\phi(z)\,dz - \int^{x+y}_{0}\phi(z)\,dz\right) = \frac{1}{t}\int^{x+y + th + tk}_{x+y}\phi(z)\,dz.$$

And I don't really know what the limit is equal to when $t \rightarrow 0$.

Answer 1

We can calculate its partials:

$$\frac{\partial}{\partial x}f(x,y)=\phi(x+y)\hspace{1cm}\text{ and}\hspace{1cm}\frac{\partial}{\partial y}f(x,y)=\phi(x+y).$$ Then along the vector $(h,k)$, the derivative is $(h\phi(x+y),k\phi(x+y))$.

Answer 2

Consider the auxiliary function $$g(u):=\int_0^u \phi(t)\>dt\ .$$ Then $g'(u)=\phi(u)$; furthermore $f(x,y):=g(x+y)$, by definition. By the chain rule we therefore have $$f_x(x,y)=g'(x+y)\cdot1=\phi(x+y),\quad f_y(x,y)=g'(x+y)\cdot1=\phi(x+y)\ .$$ It follows that $$df(x,y)=f_xdx+f_y dy=\phi(x,y)(dx+dy)\ .$$This same information can be encoded in various other ways: $$\nabla f(x,y)=\bigl(\phi(x,y),\phi(x,y)\bigr),\qquad \bigl[df(x,y)\bigr]=\bigr[\phi(x,y)\ \ \phi(x,y)\bigr]\ .$$