I am trying to find the derivative of the following function using the definition of a derivative: $$f(v)=\frac{v+4}{\lvert v+4\rvert}$$ I've already tried substituting $|v+4|$ with $\sqrt{(v+4)^2}$: $$DQ=\lim_{h \to 0} \frac{\frac{(v+h)+4}{\sqrt{((v+h)+4)^2}}-\frac{(v+4)}{\sqrt{(v+4)^2}}}{h}$$ $$DQ=\lim_{h \to 0}\frac{\frac{(v+h+4)(\sqrt{(v+4)^2}) -(v+4)(\sqrt{((v+h)+4)^2})}{\sqrt{((v+h)+4)^2}*\sqrt{(v+4)^2}}}{h}$$
$$DQ=\lim_{h \to 0} \frac{(v\sqrt{(v+4)^2} + h\sqrt{(v+4)^2} + 4\sqrt{(v+4)^2} - v\sqrt{((v+h)+4)^2} -4\sqrt{((v+h)+4)^2})}{\sqrt{((v+h)+4)^2}*\sqrt{(v+4)^2}*h}$$ , but I can't seem to simplify it enough to the point where I can cancel out the $h$.
I do know that as h approaches 0, $\sqrt{(v+4+h)}$ approaches $\sqrt{v+4}$, so I thought that maybe I could simplify some of the square roots that way
$$DQ=\lim_{h \to 0} \frac{(v\sqrt{(v+4)^2} + h\sqrt{(v+4)^2} + 4\sqrt{(v+4)^2} - v\sqrt{(v+4)^2} -4\sqrt{(v+4)^2})}{\sqrt{(v+4)^2}*\sqrt{(v+4)^2}*h}$$
$$DQ=\lim_{h \to 0} \frac{(v\sqrt{(v+4)^2} - v\sqrt{(v+4)^2}+ 4\sqrt{(v+4)^2} -4\sqrt{(v+4)^2} + h\sqrt{(v+4)^2} )}{(v+4)^2*h}$$
$$DQ=\lim_{h \to 0} \frac{h\sqrt{(v+4)^2}}{(v+4)^2*h}$$
$$DQ=\frac{\sqrt{(v+4)^2}}{(v+4)^2}$$
$$DQ=\frac{|v+4|}{(v+4)^2}$$
, but it's also kinda fishy to me since I haven't cancelled out h yet.
Any thoughts?
Solution:
If a function is differentiable at $x=x_0$, then it is continuous at $x=x_0$. Now if your function was differentiable at $v=-4$ then it would imply it is also continuous. But we know the function is not continuous at $v=-4$, since left limit and right limit are different at $v=-4$.
Note that the left limit is -1 and the right limit is +1.
But if you are solely concerned "How to cancel out h in the limit"
Just take the case where $v>-4$, then you can remove the absolute value. Then do the same for $v<-4$