# Show function g(x)=(f(x))^(1/3) is derivable, f(x)=x-sin x? what are problems whith g'(x) in 0?

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### 1 Answer

The function `g(x)` is continuous, hence, it is differentiable.

You need to test if the re exists derivative of the function `g(x)` at x = 0, using definition of derivative, such that:

`g'(0) = lim_(x->0)(g(x) - g(0))/x`

`g'(0) = lim_(x->0)(root(3)(x - sin x) - root(3)(0 - sin 0))/x`

`g'(0) = lim_(x->0)(root(3)(x - sin x))/x`

`g'(0) = root(3)(lim_(x->0)((x - sin x))/x^3) = oo/oo`

The indetermination oo/oo requests the use of l'Hospital's theorem, such that:

`g'(0) = root(3)(lim_(x->0)((x - sin x)')/((x^3)'))`

`g'(0) = root(3)(lim_(x->0)(1 - cos x)/(3x^2)) = oo/oo`

Differentiating again, yields:

`g'(0) = root(3)(lim_(x->0)(1 + sin x)/(6x)) = oo/oo`

Differentiating again, yields:

`g'(0) = root(3)(lim_(x->0)(cos x)/6) = 1/(root(3)6)`

**Hence, since derivative of the function `g(x)` , at `x = 0` , exists, yields that the function `g(x)` is differentiable at all x values.**

### Hide Replies ▲

You have done an useless step and went wrong!

`lim_(x->0) (x-sin x)/x^3= lim _(x->0)(1-cosx)/(3x^2)` (right, first form is `0/0)`

`lim_(x->0) (1-cos x)/(3x^2)=lim_(x->0) (sinx)/(6x)=1/6 lim_(x->0) sinx/x=1/6`

How did it come:

`lim_(x->0) (1-cosx)/(3x^2)= lim_(x->0) (1+sinx)/(6x)` ?!!!!!!!!!

Its a indeterminate form not `0/0` and neither comparing with a finte limit!