# Differentiability of function involving absolute values

I need to check the differentiability at $x=-10$ of the following function.

$f(x)=\cos|x-5|+\sin|x-3|+|x+10|^3-(|x|+4)^2$

Now,

\begin{align*} \text{LHD} &= \lim_{x\to -10^-}\frac{f(x)-f(-10)}{x+10}\\ &= \lim_{x\to -10^-}\frac{\cos(x-5)-\sin(x-3)-(x+10)^3-(4-x)^2-(\cos15-\sin13-14^2)}{x+10}\\ &= \lim_{x\to -10^-}\frac{2\sin(\frac{x+10}{2})\sin(\frac{20-x}{2})+2\cos(\frac{x+10}{2})\sin(\frac{16-x}{2})-(x+10)^3+(10+x)(18-x)}{x+10}\\ \\ &=-\infty \end{align*}

Similarly,

RHD = $+\infty$. So, not differentiable. But the answer says that $f(x)$ is differentiable at $-10$.

• I would add this as a comment, but I don't have enough points: the absolute value function is differentiable everywhere outside of zero, so you only need to worry there. And you know $\sin x, \cos x$ are everywhere differentiable. – RickyBobby Jul 26 '13 at 12:00
• so r u suggesting that the function is not differentiable at -10 and 0? But the answer says it is not differentiable at 3 and 0, and it is differentiable at 5 and -10. M confused. – Ramit Jul 26 '13 at 12:07
• you got the sign of sin 13 wrong when you substituted x = -10 into f(-10), it should be (cos 15+sin 13-14^2), so you should get a finite number instead of infinity. – Hugo Jul 26 '13 at 12:35
• @LongMai- thanks a ton, genius! – Ramit Jul 26 '13 at 12:47
• sure, sin(x) is differentiable everywhere, but sin(|x|) is not. – Ramit Jul 26 '13 at 16:52

Let's concentrate on the $|x+10|^3$ component since near $x=-10$ we have $$|x-5| =5-x$$$$|x-3|=3-x$$ and $$|x|+4=-x+4$$ and the components involving these quantities are readily seen to be differentiable.

Now $|x+10|$ is continuous at $x=-10$ but not differentiable there. We have

$$|x+10|^3= -(10+x)^3 \text{ for }x \le -10$$ and (noting that $x=-10$ is still a point of continuity and doesn't require a special definition) $$|x+10|^3= (10+x)^3 \text{ for }x \ge -10$$

Now if we plug a small change $h$ into the definition of the derivative for $f(x)=|x+10|^3$ we are looking at $\cfrac {f(x+h)-f(x)}h$.

We note that $f(-10)=0$, and that $f(-10+h)=|h|^3$. So $$\frac{f(-10+h)-f(-10)}h=\frac {|h|^3}h=\frac {h\cdot|h|^3}{h^2}=h\cdot|h|$$

(note $h^2=|h|^2$), and you should be able to complete the problem from there.

f(x)=cos(x-5)- sin (x-3)-(x-3)^3-(-x+4)^2 when xis less than -10 or = -10

=cos(x-5)-sin(x-3)+(x-3)^3-(-x+4)^2 whenis between-10 and 0

LHD at x=-10

=sin 15- c0s 13 -196

RHD AT x=-10

=sin15-cos 13 -196

icomputed the lrft hand derivative and right hand derivative using the definition and l`hopital rule

so the derivative at x=-10 exists and the function f is differentiable at x=-10