Derivative of |sin x| where it's differentiable

[PirateFan308]

Homework Statement

Compute the derivatives of the following (where they are differentiable):
h) |sinx|

Homework Equations

Chain rule: (f°g)'(c) = f'(g(c))(g'(c))

The Attempt at a Solution

Let f=|x| and g=sin x
(f°g)'(c) = f'(g(x))g'(c) = f'(sin x)(cos x)
But I don't know what (|x|)' is. It's +1 when x>0 and -1 when x<0 and it's not differentiable at 0, but then there is no x to plug g into, and looking at the graph, I don't think this would be right. Thanks!

 

[Dickfore]

Hint:
$$\vert x \vert \equiv \left\lbrace\begin{array}{rl}<br /> x &, \ x \ge 0 \\<br /> <br /> -x &, \ x < 0<br /> \end{array}\right.$$

Therefore:
$$\frac{d}{d x} \vert x \vert = \left\lbrace \begin{array}{rl}<br /> 1 &, \ x > 0 \\<br /> <br /> -1 &, \ x < 0<br /> \end{array}\right. \equiv \mathrm{sgn}(x)$$
The derivative does not exist at $x = 0$.

Then, use the chain rule.

 

[Simon Bridge]

express the function piecewise to remove the absolute value sign.
eg |x| = {x: x>0 and -x: x<0 and 0: x=0}

[edit]Dickfore beat me to it

 

[PirateFan308]

Hint:
[tex] Therefore:<br /> $$\frac{d}{d x} \vert x \vert = \left\lbrace \begin{array}{rl}<br /> 1 &, \ x > 0 \\<br /> <br /> -1 &, \ x < 0<br /> \end{array}\right. \equiv \mathrm{sgn}(x)$$<br /> The derivative does not exist at $x = 0$.<br /> <br /> Then, use the chain rule.[/tex]

[tex] <br /> So if x>0, $(f°g)' = f'(g(x))g'(x) = (1)(cos x) = cos x$<br /> <br /> If x<0, $(f°g)' = f'(g(x))g'(x) = (-1)(cos x) = -cos x$<br /> <br /> Thanks for the hint![/tex]

 

[SammyS]

So if x>0, $(f°g)' = f'(g(x))g'(x) = (1)(cos x) = cos x$

If x<0, $(f°g)' = f'(g(x))g'(x) = (-1)(cos x) = -cos x$

Thanks for the hint!

What you have is not correct. sin(x) may be negative when x > 0 .

Your function is:

$\vert \sin(x) \vert \equiv \left\lbrace\begin{array}{rl} <br /> \sin(x) &\text{if } \ \sin(x) \ge 0 \\ <br /> <br /> -\sin(x) &\text{if } \ \sin(x) < 0 <br /> \end{array}\right.$

So, what matters is the sign of sin(x), not the sign of x itself.

 

[ehild]

You can use the "sign" function sign(a)=a/|a| (http://mathworld.wolfram.com/Sign.html) to express the derivative of |sinx|.

It is also possible to use the identity

$|a|=\sqrt{a^2}$

and determine the derivative of $\sqrt{sin^2(x)}$.

ehild

 

[PirateFan308]

Thanks!