Apply dominated convergence theorem to show differentiability Let $f,g \in L^p(\mu), 1 < p < \infty$. Show that the function
$$\phi(t)=\int |f+tg|^pd\mu$$ be differentiable at $t=0$ and find $\phi'(0)$.

My try, 
$\psi(t)=|f+tg|^p$ is differentiable at $t=0$, if $p>1$. So
$$\lim_{t\to 0}\frac{|f+tg|^p-|f|^p}{t}=h(x)$$
I also showed 
$$t\in(0,1),\frac{|f+tg|^p-|f|^p}{t}\leq |f+g|^p-|f|^p$$
$$t\in(-1,0),\frac{|f+tg|^p-|f|^p}{-t}\leq |f-g|^p-|f|^p$$
Then I think I need to use dominated convergence theorem to show
$$\lim_{t\to 0}\int\frac{|f+tg|^p-|f|^p}{t}d\mu=\int h(x)d\mu$$
But what's a suitable dominator? I think I only found an integrable upper bound for the function, not the absolute value of the function.

Hint: $\big||f+tg|^p-|f|^p\big| = p|tg|\cdot|\xi|^{p-1} $, where (for any $x$) $\min\{f+tg,f\}\leq \xi \leq \max \{f+tg,f\}$. Hence (wlog assume $t<1$, since you are really interested in $t\to 0$) $|\xi|\leq |f|+|g|$, i.e.
$$\big||f+tg|^p-|f|^p\big| = p|tg|\cdot|\xi|^{p-1}\leq p t |g|(|f|+|g|)^{p-1}\leq pt|g|\cdot \max \{2, 2^{p-2}\}(|f|^{p-1}+|g|^{p-1})$$
and Holder to control $\int |f|^{p-1}|g|$...
