Derivative of x^2sin(1/x)

Answer: The derivative of f(x) defined by f(x) = x²sin(1/x) if x≠0 and 0 if x=0 is equal to 0. That if, if

f(x) = $\begin{cases} x^2 \sin \left(\dfrac{1}{x} \right) & \text{ if } x \neq 0 \\ 0 & \text{ if } x=0 \end{cases}$,

then $f'(0)=0$.

Differentiate x^2sin(1/x)

Question: Find $f'(0)$ where f(x) is defined as f(x) = $\begin{cases} x^2 \sin \left(\dfrac{1}{x} \right) & \text{ if } x \neq 0 \\ 0 & \text{ if } x=0. \end{cases}$

Answer:

From the definition of derivatives, the first order derivative of f(x) at x=0 is given by

$f'(0)$ = lim_h→0 $\dfrac{f(0+h)-f(0)}{h}$

= lim_h→0 $\dfrac{f(h)-f(0)}{h}$

= lim_h→0 $\dfrac{h^2 \sin \left(\dfrac{1}{h} \right)-0}{h}$

⇒ $f'(0)$ = lim_h→0 $h \sin \left(\dfrac{1}{h} \right)$ …(∗)

Now, to compute the limit of h sin(1/h) as h tends to 0, we will use the sandwich/squeeze theorem on limits.

For $h>0$, we have $-h \leq h \sin \left(\dfrac{1}{h} \right) \leq h$ …(1)

For $h<0$, we have $h \leq h \sin \left(\dfrac{1}{h} \right) \leq -h$ …(2)

Combining (1) and (2), it follows that

$-|h| \leq h \sin \left(\dfrac{1}{h} \right) \leq |h|$

Now, taking lim_h→0 on both sides, we obtain that

$\lim\limits_{h \to 0}-|h| \leq$ $\lim\limits_{h \to 0}h \sin \left(\dfrac{1}{h} \right) \leq$ $\lim\limits_{h \to 0}|h|$

⇒ $0 \leq$ lim_h→0 $h \sin \left(\dfrac{1}{h} \right) \leq 0$

⇒ $h \sin \left(\dfrac{1}{h} \right) = 0$

Therefore, from (∗), we get that $f'(0) =0$. So the derivative of x²sin(1/x) at x=0 is equal to 0.

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