- #1

MatinSAR

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- 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.

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.