Question

How to find `f'(0)` where `f(x)=root(5)(x^3-tan^3x)`?

Answer 1

`f'(0)=0`

Explanation:

As `f(x)=root(5)(x^3-tan^3x)=(x^3-tan^3x)^(1/5)`

hence `(df)/(dx)=1/5(x^3-tan^3x)^(-4/5)xx(3x^2-3tan^2xsec^2x)`

and `f'(0)=0` `->` as `0^3-tan^3 0=0`

Answer 2

See below.

Explanation:

`lim_(x->0)d/(dx)root(5)(x^3 - tan^3x)`

`d/(dx)root(5)(x^3 - tan^3x)=(3 (x^2 - sec^2xTan^2x))/(5 (x^3 - tan^3x)^(4/5))`

and for small `abs(x)` we have

`(3 (x^2 - sec^2xTan^2x))/(5 (x^3 - tan^3x)^(4/5)) approx 3/5(x^2-sin^2x)/(x^3-sin^3x)^(4/5)=3/5((x+sinx)(x-sinx))/((x-sinx)(x^2+xsinx+sin^2x))^(4/5) =`

`(3/5)((x^2)/(x^3)^(4/5))((1+sinx/x)(1-sinx/x))/((1-sinx/x)(1+sinx/x+sin^2x/x^2))^(4/5)`

and for small `abs(x)` we have

`(3/5)((x^2)/(x^3)^(4/5))((1+sinx/x)(1-sinx/x))/((1-sinx/x)(1+sinx/x+sin^2x/x^2) )^(4/5)approx (3/5)1/x^(2/5)(2(1-sinx/x))/((1-sinx/x)^(4/5)3^(4/5))=2(3/5)3^(-4/5)((1-sinx/x)^(1/5)/x^(2/5))=`
`=2(3/5)3^(-4/5)((1-sinx/x)/x^2)^(1/5)`

Now

`lim_(x->0)d/(dx)root(5)(x^3 - tan^3x)=lim_(x->0)2(3/5)3^(-4/5)((1-sinx/x)/x^2)^(1/5)=2(3/5)3^(-4/5)(1/6)^(1/5)=2^(4/5)/5`

NOTE:

`(1-sinx/x)/x^2 approx (1- (x-x^3/(3!)+ cdots)/x)/x^2`