Show that f has a unique zero and prove that f′(λ)>0 with f being continuous and differentiable. Let $f: [a,b] \longrightarrow \mathbb{R}( a<b)$, $f$ is continuous and differentiable. 
We assume that $f$ and $f'$ are increasing and $f(a)<0, 0<f(b)$.
Show that $f$ has a unique zero which we denote $\lambda$ and prove that $f'(\lambda)>0$
I have used IVT theorem to show that $\lambda$ exists.
*For the uniqueness if $f'$ is continuous, I can use 
 $$f(\mu)-f(\lambda)=\int_\lambda^\mu f'(x) \, \mathrm dx$$
Except that f is only differentiable..
Thank you in advance for your help

You don't need the fact that $f$ is increasing, only $f'$.
First, as you said, use IVT for existence.
Second, consider your integral $f(a)-f(0)$ where $f(a)=0$ and conclude that $f'(a)$ must be positive. (Edit: as per your comment, if you don't like integral, use MVT to show the same fact.)
Finally, given this, conclude that for $b>a$: $f(b)=f(b)-f(a)$ must be positive as well.
