**Toolbox:**

- If a function is differentiable in an interval it is continuous in that interval.
- As per L.M.V Theorem if a function is continuous and differentiable then $ \exists $ atleast point $ c \in ( a, b )$ such that $ f'(c) = \large\frac{f(b)-f(a)}{b-a}$

Step 1

From L.M.V theorem

$ f' (c)=\large\frac{10-4}{b-a}\: and \: g' (c)=\large\frac{3-1}{b-a}$

$ f' (c) =\large \frac{6}{b-a} \: and \: \large\frac{2}{b-a} = g' (c)$

Step 2

$ \Rightarrow f' (c) = 3 g' (c)$