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: 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.
A: Assume $f(\lambda)=0$ (you have already show the existence of at least one such $\lambda$ with the IVT). Then by the MVT,
$$f(\lambda)-f(a)=(\lambda-a)f'(\xi)$$
for some $\xi$ with $a<\xi<\lambda$. As $f(\lambda)-f(a) = -f(a) >0$ and $\lambda-a\ge 0$ we conclude (because $f'$ is increasing) that
$$ f'(\lambda)\ge f'(\xi)>0.$$
Assume  $\lambda<\mu$ with $f(\lambda)=f(\mu)=0$. Them the MVT leads (again using that $f'$ is increasing) to the contradiction
$$ f(\mu)-f(\lambda)=(\lambda-\mu)f'(\xi)\ge (\lambda-\mu)f'(\lambda)>0.$$
Remark: We did not use that $f$ is increasing.
A: 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 $a<x<\lambda$.
