f(x) = 3√ x · sin x use the definition of derivative to show f(x) is differentiable at x=0

Question

this question had several parts 

1)Why can't we use the product rule to compute f'(0) 

 

My initial thought was that you cannot use the product rule t because sinx and 3√ x both equal 0 at x=0 

 

2)use the definition of derivative to show f(x) is differentiable at x=0

does this mean to take the limit of of f(x+h)-f(x)/h  as h > 0- and h>0+  

how would i approach this problem

 

Answer 1

f(x) = 3 √x - sinx   (note that x ≥ 0)

 

f'(0) = lim_h→0 [ (f(0+h) - f(0))/h ]

 

       = lim_h→0 [ (3√h - sinh) / h ]

 

       = lim_h→0 [3/√h - sinh/h] = ∞ - 1 = ∞

 

The given function is NOT differentiable when x = 0.  The function has a vertical tangent at the point (0,0).