Help to see if this function is diffentiable (1 Viewer)

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Hello,
I need some help to see if this function is diffentiable at [tex]x_{0}=\frac{\pi}{4}[/tex]
the function is f(x)=|sin(x)-cos(x)|
To do that, you have to find the limit of [tex]\frac{f(x)-f(x_{0})}{x-x_{0}}[/tex] as x-> pi/4
So I get [tex]\frac{sin(x)-cos(x)}{x-\frac{\pi}{4}}[/tex]but I don't know what to do after...any help or hints would be appreciated
Thanks
 
Last edited:

benorin

Homework Helper
1,057
7
since sin(x) and cos(x) are equal at x=pi/4, we have

[tex]f(x) = | \sin (x)- \cos (x)| = \left\{\begin{array}{cc}\cos (x)-\sin (x) ,&\mbox{ if } 0\leq x\leq \frac{\pi}{4}\\ \sin (x)- \cos (x), & \mbox{ if } \frac{\pi}{4}\leq x \leq \pi\end{array}\right. [/tex]

now compute the derivative using the formula

[tex]f^{\prime} (x_0) = \lim_{h\rightarrow 0} \frac{f(x_0+h) -f(x_0)}{h}[/tex]

since f(x) is piecewise defined, use left- and right-handed limits to comput the above limit, here is the first one

[tex]f_{-}^{\prime} \left( \frac{\pi}{4}\right) = \lim_{h\rightarrow 0^{-}} \frac{\cos\left( \frac{\pi}{4}+h\right) -\sin\left( \frac{\pi}{4}+h\right) -0}{h} = \lim_{h\rightarrow 0^{-}} \frac{\cos\left( \frac{\pi}{4}\right) \cos (h) -\sin\left( \frac{\pi}{4}\right) \sin (h) - \sin\left( \frac{\pi}{4}\right) \cos(h)- \cos\left( \frac{\pi}{4}\right) \sin (h)}{h}[/tex]
[tex] =\frac{\sqrt{2}}{2} \lim_{h\rightarrow 0^{-}} \frac{-2\sin (h)}{h} = -\sqrt{2}[/tex]

now do the right-hand limit to finish-up.
 

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