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

A different approach without need of integrals.
Since $f$ is increasing, $f'(x)\ge0$. Since $f(a)<0$ and $f(b)>0$, there is a $t\in(a,b)$ such that $f(t)=0$. Let $\lambda=\inf_{t\in(a,b)}f(t)=0$. By continuity, $f(\lambda)=0$ , and $a<\lambda<b$.  If $f'(\lambda)=0$, then $f'(x)=0$ for $x\in(a,\lambda)$. This implies that $f$ is constant in$(a,\lambda)$, a contradiction since $f(a)<0$. Thus $f'(\lambda)>0$. This implies that $f(x)>0$ if $\lambda<x<b$ and $f(x)<0$ if $a<x<\lambda$.
