We know that, #f'(x)=lim_(t to x){f(t)-f(x)}/(t-x)#.
So, if #f(x)=root(3)tanx=tan^(1/3)x#, then,
#f'(x)=lim_(t to x){tan^(1/3)t-tan^(1/3)x)/(t-x)#,
#=lim(tan^(1/3)t-tan^(1/3)x)/(tant-tanx)*(tant-tanx)/(t-x)#,
# i.e., f'(x)=L_1*L_2"............................."(ast)," say, where, "#
#L_1=lim_(t to x)(tan^(1/3)t-tan^(1/3)x)/(tant-tanx)#.
Here, let #tan^(1/3)t=T and tan^(1/3)x=X#.
#:." As "t to x, T to X; and, tant=T^3, tanx=X^3#.
#:. L_1=lim_(T to X)(T-X)/(T^3-X^3)#,
#=lim(T-X)/{(T-X)(T^2+TX+X^2)}#,
#=lim_(T to X) 1/(T^2+TX+X^2)#,
#=1/(X^2+X*X+X^2)#.
# rArr L_1=1/(3X^2)=1/(3tan^(2/3)x).................(ast^1)#.
#L_2=lim_(t to x )(tant-tanx)/(t-x)#,
#=lim(sint/cost-sinx/cosx)/(t-x)#,
#=lim(sintcosx-costsinx)/{(cost)(cosx)(t-x)}#,
#=lim_(t to x){(sin(t-x))/(t-x)}sect*secx#,
#=1*secx*secx#.
# rArr L_2=sec^2x....................................(ast^2)#.
Combining #(ast), (ast^1) and (ast^2)#, we have,
#[root(3)tanx]'=sec^2x/(3tan^(2/3)x)#, as Respected Maganbhai P.
has derived!