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!