Derivative of $\int_0^1 e^{\sqrt{x^2+t^2}}\,\mathrm{d}x$ at $t = 0$ Let the real-valued function $\phi:\mathbb{R}\to\mathbb{R}$ be defined by
$$\phi(t)=\int_0^1e^{\sqrt{x^2+t^2}}\,\mathrm{d}x,$$
it can then be shown that $\phi$ is continuous and differentiable. 
I wish to find its derivative, in particular at $\mathbf{t = 0}$. I believe, since the integration is independent of $t$, that I can differentiate under the integration sign (correct me if I'm wrong) hence
$$\phi'(t)=\frac{\mathrm{d}}{\mathrm{d}t}\int_0^1e^{\sqrt{x^2+t^2}}\,\mathrm{d}x =  \int_0^1 \frac{\mathrm{d}}{\mathrm{d}t}e^{\sqrt{x^2+t^2}}\,\mathrm{d}x = \int_0^1 \frac{te^{\sqrt{x^2+t^2}}}{\sqrt{x^2+t^2}}\,\mathrm{d}x.$$
Naïvely, it then looks like $\phi'(0) = 0$, since it is $0$ times something. However, as $t\to0$ we actually get the indeterminate form $0\cdot \infty$ since the integrand becomes infinite, so we cannot say that $\phi'(0) = 0$.
I tried various substitutions, but I'm not sure if this is the right approach. Perhaps, it is necessary to use dominated convergence theorem or something of that sort?
Thanks for any help.

Your last integral equals 
$$\int_0^1 \frac{e^{\sqrt{x^2+t^2}}}{\sqrt{(x/t)^2+1}}\,dx.$$
The integrands are uniformly bounded for small $t.$ They converge pointwise to $0.$ Hence $\phi'(t)\to 0$ as $t\to 0$ by the dominated convergence theorem. This implies $\phi'(0) = 0.$
