# Differential calculus of Functions

• Nov 22nd 2010, 12:29 AM
ZOOZ
Differential calculus of Functions
i got 2 Functions Y(x)=lxl^3 f(x)=x-1

1) Prove that Y can be Differentiable In point 0

2) use the
information from ur Proof In 1 to Prove that the Function l(x-1)l^3
are
Differentiable In point 1

thanks for helping.
• Nov 22nd 2010, 02:30 AM
DrSteve
$\lim_{h\rightarrow 0}\frac{Y(0+h)-Y(0)}{h}=\lim_{h\rightarrow 0}\frac{|h|^3}{h}$.

Now compute the limit from the left and the limit from the right separately. They both give 0, showing that the derivative of Y at 0 exists and is equal to 0.

For the second one use the derivative theorem about the composition of 2 functions.

If you need clarification on the right and left hand limits let me know.
• Nov 22nd 2010, 03:06 AM
ZOOZ
but i need clarification on the right and left hand limits.

U mean from left lim -0 is still give 0
like lim 0+ is also give 0
?

thnaks
• Nov 22nd 2010, 03:19 AM
Plato
Quote:

Originally Posted by ZOOZ
but i need clarification on the right and left hand limits. U mean from left lim -0 is still give 0
like lim 0+ is also give 0

$h < 0\, \Rightarrow \,\frac{{\left| h \right|^3 }}
{h} = - h^2$
• Nov 22nd 2010, 03:19 AM
DrSteve
Yes. If h is positive, then |h| is just h. If h is negative, then |h| is -h.

So if h is positive $\frac{|h|^3}{h} = \frac{h^3}{h} = h^2$. So the limit from the right is $0^2 = 0$.

If h is negative, $\frac{|h|^3}{h} = \frac{(-h)^3}{h} = -h^2$. So the limit from the left is $-0^2 = 0$.
• Nov 22nd 2010, 03:23 AM
ZOOZ
OK thanks so much i will try alone to Solve the 2) Question.