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

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.
