Solow Model - speed of convergence

This is a question also for those with a good expertise in micro. For micro guys who wanna go streight to the question, just jump to equation $$(1)$$

I'm studying the Solow growth model.

Let's write the law of motion of capital in intensive form (with technology normalized to one):

$$\dot{k}(t)=s f(k(t)) -(\delta + n)k(t) \equiv \phi (k(t))$$

where $$k(t) \equiv \frac{K(t)}{L(t)}$$, $$f(k(t))=\frac{F(K(t),L(t),1)}{L(t)}$$,

$$\delta$$ denotes the depreciation rate of capital, and $$n$$ is the population growth rate.

Then, taking a linear approximation of $$\phi (k)$$ around $$k^\ast$$

$$\dot{k}(t) \approx \ \phi(k^\ast) + \phi'(k^\ast) (k(t)-\phi(k^\ast))= \phi'(k^\ast) (k(t)-\phi(k^\ast))$$

Solving this linear differential equation, we get

$$k(t) = k^\ast + e^{-\lambda t} (k(0)-k^\ast)$$, with $$\lambda \equiv - \phi'(k^\ast) >0$$

The speed of convergence is determined by $$\lambda$$

$$\lambda \equiv - \phi'(k^\ast) = - \frac{ s f'(k^\ast)k^\ast}{k^\ast} + (\delta +n)$$

Which boils down into

$$\lambda= (1- \frac{f'(k^\ast)k^\ast}{f(k^\ast)})(\delta +n) >0$$ $$~~~~~~(1)$$

The slides says:

The speed of convergence determined by (i) the degree of concavity in the production function and (ii) the effective depreciation rate.

Can you tell me why $$\frac{f'(k^\ast)k^\ast}{f(k^\ast)}$$ describes the degree of concavity of the production function ?