# Derivative of a piecewise function

• MatinSAR
In summary, the function given is a plain old x^2 which has a thin disguise of being a function at x=1.
MatinSAR
Homework Statement
Finding derivative of function ##f## which is a piecewise function.
Relevant Equations
##f'(a)=\lim_{x \rightarrow a} {\frac {f(x)-f(a)} {x-a}}##
##f(x) = \begin{cases} \frac {x^3-x^2}{x-1} & \text{if } x\neq1 \\ 1 & \text{if } x=1 \end{cases}##
Find ##f'(1)##.

Using derivative definition:
##f'(1)=\lim_{x \rightarrow 1} {\frac {\frac {x^3-x^2}{x-1}-f(1)} {x-1}}##

##f'(1)=\lim_{x \rightarrow 1} {\frac {\frac {x^3-x^2}{x-1}-1} {x-1}}##

##f'(1)=\lim_{x \rightarrow 1} {\frac {x^3-x^2-x+1} {(x-1)^2}}##

Apply L'Hôpital's rule two times:
##f'(1)=\lim_{x \rightarrow 1} {\frac {6x-2} {2}}##
##f'(1)=2##

Can someone please tell me if I'm wrong …

Can I say that ##f(x) = \begin{cases} \frac {x^3-x^2}{x-1} & \text{if } x\neq1 \\ 1 & \text{if } x=1 \end{cases}## graph is similar to ##y=x^2## but it's undefined at ##x=1##? If it's true then I can find ##f'(1)=2x=2## far faster.

##f(x) = x^2##? True or false?

MatinSAR
PeroK said:
##f(x) = x^2##? True or false?
No these functions have different domains, so they're not equal.

MatinSAR said:
No these functions have different domains, so they're not equal.
The domain is ##\mathbb R## in both cases.

It's true. The function you are given is plain old ##x^2##, wearing a thin disguise!

MatinSAR
PeroK said:
The domain is ##\mathbb R## in both cases.

It's true. The function you are given is plain old ##x^2##, wearing a thin disguise!
Oh! I've forgotten that we've defined ##f(1)=1##. So both ways were correct, Do you agree?

MatinSAR said:
Oh! I've forgotten that we've defined ##f(1)=1##. So both ways were correct, Do you agree?
Both methods are valid. But, the first method is unnecessarily complicated.

As an aside, I've never liked the term piecewise function, as there is no such thing, IMO. You can define a function piecewise. But, as in this case, that doesn't mean that you have to define it piecewise. In this case, you could equally well have written ##f(x) = x^2 \ (\forall x)##.

I wonder whether whoever set the question realised this.

MatinSAR
PeroK said:
Both methods are valid. But, the first method is unnecessarily complicated.
Yes.
PeroK said:
As an aside, I've never liked the term piecewise function, as there is no such thing, IMO. You can define a function piecewise. But, as in this case, that doesn't mean that you have to define it piecewise. In this case, you could equally well have written ##f(x) = x^2 \ (\forall x)##.

I wonder whether whoever set the question realised this.
I've understand it. Thank you for your valuable help and time.

## 1. What is a piecewise function?

A piecewise function is a function that is defined by different expressions or formulas over different intervals of its domain. Each piece of the function applies to a specific interval, and the function can have different behaviors in each interval.

## 2. How do you find the derivative of a piecewise function?

To find the derivative of a piecewise function, you need to differentiate each piece of the function separately within its respective interval. After differentiating, you must check the continuity and differentiability at the points where the pieces meet (the boundaries) to ensure the overall function is differentiable at those points.

## 3. What are the conditions for a piecewise function to be differentiable at a boundary point?

For a piecewise function to be differentiable at a boundary point, two conditions must be met: 1) The function must be continuous at the boundary point.2) The left-hand derivative and the right-hand derivative at that point must be equal. If both conditions are satisfied, the function is differentiable at the boundary point.

## 4. How do you check the continuity of a piecewise function at a boundary point?

To check the continuity of a piecewise function at a boundary point, you need to ensure that the limit of the function as it approaches the boundary point from the left is equal to the limit as it approaches from the right, and both of these limits must equal the value of the function at that point. Mathematically, this means that $$\lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x) = f(c)$$.

## 5. Can a piecewise function be continuous but not differentiable at a boundary point? Why?

Yes, a piecewise function can be continuous but not differentiable at a boundary point. This occurs when the left-hand derivative and the right-hand derivative at the boundary point are not equal. While the function may be continuous (no jumps or breaks), the difference in the slopes (derivatives) from the left and right means that the function has a sharp corner or cusp at that point, making it non-differentiable there.

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