1. ## Derivative help

Hello, so I have this problem that I am not completely sure on, here it is:

Let f be the real-valued function defined by $f(x)=sin^3x + sin^3|x|$

(a) Find $f'(x) for x>0.$
(b) Find $f'(x) for x<0.$
(c) Determine whether $f(x)$ is continuous at $x=0$. Justify your answer
(d) Determine whether the derivative of $f(x)$ exists at $x=0$. Justify your answer.

So I know how to do power rule and I have memorized some standards derivative values. But I do not know about raising sin to the third power. If $x$ was cubed I would be able to do it, but not sin.

For b it would be different because the absolute value of x $= \sqrt(x^2)$. I do not know how to take the derivative of that.

For c I said no because sin is never continuous at 0?

Finally for d i out that it does not because $f(x)$ is not continuous at 0.

ANY help would be greatly appreciated once again. Thanks.

2. Originally Posted by OnMyWayToBeAMathProffesor
So I know how to do power rule and I have memorized some standards derivative values. But I do not know about raising sin to the third power. If $x$ was cubed I would be able to do it, but not sin.
Are you familiar with the chain rule? You will need to use that here.

Let $f$ be the real-valued function defined by $f(x)=\sin^3x+\sin^3\lvert x\rvert$

(a) Find $f'(x)\text{ for }x>0.$
For $x>0,\;\lvert x\rvert=x.$ So,

$f'(x)=\frac d{dx}\left[\sin^3x+\sin^3x\right]$

$=\frac d{dx}\left[2\sin^3x\right]$

$=\left(6\sin^2x\right)\frac d{dx}\left[\sin x\right]$

$=6\sin^2x\cos x.$

(b) Find $f'(x)\text{ for }x<0.$
When $x<0,\;\lvert x\rvert=-x.$ So, remembering that sine is an odd function ( $\sin(-a)=-\sin a$), we have

$f'(x)=\frac d{dx}\left[\sin^3x+\sin^3(-x)\right]$

$=\frac d{dx}\left[\sin^3x+(-\sin x)^3\right]$

$=\frac d{dx}\left[\sin^3x-\sin^3x\right]$

$=0.$

(c) Determine whether $f(x)$ is continuous at $x=0$. Justify your answer
...
For c I said no because sin is never continuous at 0?
Why would you think that? Sine is continuous over the entire real line.

$\lvert x\rvert$ is everywhere continuous, $\sin x$ is everywhere continuous, and $x^3$ is everywhere continuous. The composition of two continuous functions is itself continuous, so $\sin^3x$ and $\sin^3\lvert x\rvert$ are both continuous. Finally, the sum of two continuous functions is continuous, so $f(x)$ is everywhere continuous (including $x=0$).

If you like, you could also use the definition of continuity at a point by showing that $\lim_{x\to0}f(x)=f(0)$ (to do this, consider the one-sided limits).

(d) Determine whether the derivative of $f(x)$ exists at $x=0$. Justify your answer.
...
Finally for d i out that it does not because $f(x)$ is not continuous at 0.
See above. $f$ is continuous at $x=0,$ and in fact, $f$ is also differentiable at $x=0.$ Consider the left- and right-sided limits.

$\lim_{x\to0^-}\frac{f(x)-f(0)}x$

$=\lim_{x\to0^-}0=0$

and

$\lim_{x\to0^+}\frac{f(x)-f(0)}x$

$\lim_{x\to0^+}6\sin^2x\cos x=6\sin^20\cos0=0.$

Both one-sided limits are equal, so the limit $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}x$ exists.

3. I am sorry i was thinking of $\frac{sinx}{x}$, which is not continuous at 0. and how does $|x|=-x for x<0$ ? I thought if $x=-6$ which would be $|-6|$ which is $6$, correct?

4. nvm, i see what happened. thank you very much for a clear and concise explanation of the problem and the solution.