Derivative of |sinx|: Solving Homework Problem

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

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

<br /> <br /> So if x&gt;0, (f°g)&amp;#039; = f&amp;#039;(g(x))g&amp;#039;(x) = (1)(cos x) = cos x<br /> <br /> If x&lt;0, (f°g)&amp;#039; = f&amp;#039;(g(x))g&amp;#039;(x) = (-1)(cos x) = -cos x<br /> <br /> Thanks for the hint!

 

[SammyS]

So if x>0, (f°g)&#039; = f&#039;(g(x))g&#039;(x) = (1)(cos x) = cos x

If x<0, (f°g)&#039; = f&#039;(g(x))g&#039;(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) &amp;\text{if } \ \sin(x) \ge 0 \\ <br /> <br /> -\sin(x) &amp;\text{if } \ \sin(x) &lt; 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!