derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition

When not using the derivative definition I get $\cos (1/x) + 2x \sin(1/x)$, which WolframAlpha agrees to.

However when I try solving it using the derivative definition: $$\lim_ {h\to 0} = \frac{f(x+h) - f(x)}{h} $$

I get: $$2x \sin \left(\frac{1}{x+h} \right ) + h \sin \left(\frac{1}{x+h}\right)$$

which in return results in, as $h \to 0$:

$$2x \sin (1/x)$$

So what am I doing wrong when using the def of derivatives?

  • 3
    $\begingroup$ I am reasonably sure that you are solving the wrong problem. Probably you are told that $f(x)=x^2\sin(1/x)$ when $x\ne 0$, and $f(0)=0$. You are then asked to use the definition of derivative to find $f'(0)$, nothing else. $\endgroup$ – André Nicolas Sep 21 '14 at 14:52




where $u(x),v(x)$ are functions of $x$

Setting $v(x)=\sin\dfrac1x,$


Using Prosthaphaeresis Formula,this becomes


$$=-2\dfrac{\sin\dfrac h{2x(x+h)}}{\dfrac h{2x(x+h)}}\cdot\frac1{2x(x+h)}\cdot\cos\dfrac{2x+h}{2x(x+h)}$$


Typically, this problem is not presented quite how you've described. Instead, for $x \neq 0$ we can compute the derivative of your function with the product and chain rules. For $x=0$ of course your function is not defined, but because $\sin$ is bounded, it can be continuously extended by defining $f(0)=0$.

To find the derivative at zero, you can use the definition. This is less terrible than it would be in the general case, because your difference quotient is

$$\frac{h^2 \sin(1/h) - 0}{h} = h \sin(1/h)$$

for which the limit is straightforward to compute.


You're doing wrongly the computation: \begin{align} f(x+h)-f(x)&=(x+h)^2\sin\frac{1}{x+h}-x^2\sin\frac{1}{x}\\ &=x^2\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)+2hx\sin\frac{1}{x+h}+h^2\sin\frac{1}{x+h} \end{align} When you divide by $h$ you get $$ \frac{f(x+h)-f(x)}{h}= \frac{x^2}{h}\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)+2x\sin\frac{1}{x+h}+h\sin\frac{1}{x+h} $$ and you're left with computing $$ \lim_{h\to0}\frac{1}{h}\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right) $$ because the second summand above tends to $2x\sin(1/x)$ and the third summand tends to $0$. You'll reinsert $x^2$ later.

Now you can use the identity $$ \sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} $$ Setting $\alpha=1/(x+h)$ and $\beta=1/x$, we have $$ \frac{\alpha+\beta}{2}=\frac{1}{2}\left(\frac{1}{x+h}+\frac{1}{x}\right)= \frac{2x+h}{2x(x+h)} $$ while $$ \frac{\alpha-\beta}{2}=\frac{1}{2}\left(\frac{1}{x+h}-\frac{1}{x}\right)= \frac{-h}{2x(x+h)} $$ so your limit is $$ \lim_{h\to0}\frac{2}{h}\cos\frac{2x+h}{2x(x+h)}\sin\frac{-h}{2x(x+h)} $$ But $$ \lim_{h\to0}2\cos\frac{2x+h}{2x(x+h)}=2\cos\frac{1}{x} $$ so we just need to compute $$ \lim_{h\to0}\frac{1}{h}\sin\frac{-h}{2x(x+h)} $$ Set $k=-h/(2x(x+h))$, so $$ 2x^2k+2xhk=-h $$ or $h(2xk+1)=-2x^2k$ that lends $$ h=-\frac{2x^2k}{2xk+1} $$ So this transformation is bijective (and bicontinuous) and $h\to0$ implies $k\to0$ so your limit is $$ \lim_{h\to0}\frac{1}{h}\sin\frac{-h}{2x(x+h)}= \lim_{k\to0}-\frac{2xk+1}{2x^2k}\sin k= \lim_{k\to0}-\frac{2xk+1}{2x^2}\frac{\sin k}{k}=-\frac{1}{2x^2} $$ In the end we get $$ \lim_{h\to0}\frac{f(x+h)-f(x)}{h}= x^2\left(-\frac{1}{2x^2}\right)\cos\frac{1}{x}+2x\sin\frac{1}{x}= -\cos\frac{1}{x}+2x\sin\frac{1}{x} $$

This assumes, of course, $x\ne0$. If your function is defined by $$ f(x)=\begin{cases} x^2\sin\dfrac{1}{x}&\text{for $x\ne0$}\\[1ex] 0&\text{for $x=0$} \end{cases} $$ we are left with the derivative at $0$, that is $$ \lim_{h\to0}\frac{f(h)-f(0)}{h}= \lim_{h\to0}\frac{1}{h}h^2\sin\frac{1}{h}= \lim_{h\to0}h\sin\frac{1}{h}=0 $$ with an easy application of the squeeze theorem.

Of course using the chain rule is much simpler.

Now that you know that $$ f'(x)=\begin{cases} -\cos\dfrac{1}{x}+2x\sin\dfrac{1}{x}&\text{for $x\ne0$}\\[1ex] 0&\text{for $x=0$} \end{cases} $$ it should be easy to verify whether the derivative is continuous at $0$.

  • 1
    $\begingroup$ I was wondering what happens to the $h $ variable in the denominator of $\frac {2}{h} $ when we're computing $\lim_{h\to 0}\frac {2}{h}\cos\frac {2x+h}{2x (x+h)}\sin\frac {-h}{2x (x+h)} $? Also, I wanted to know if you could further explain why the said transformation is bijective and bicontinuous? And why is that important for the given problem? $\endgroup$ – K.M Nov 4 '18 at 19:09
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    $\begingroup$ @K.M That's taken care of when I compute the limit with the sine involved. The substitution is surely bicontinuous (where necessary): just compute its inverse. $\endgroup$ – egreg Nov 4 '18 at 20:57
  • $\begingroup$ By assumption, since we're computing the derivative of $x^2\sin\frac{1}{x}$, we only need to take into consideration non-zero $x$. So since we're computing the derivative at this non-zero $x$, I'm guessing we can sort of treat it like it's a constant? In doing so, I do get that the function $k(h)$ is both injective and surjective (bijective). And I do get that both $k$ and $h$ are continuous for relatively small $k$ and $h$, depending on the context. But I don't think that that really matters in the problem since $\lim_{k\to 0}\frac {\sin k}{k}=1$ so we no longer have $k$ in the denominator. $\endgroup$ – K.M Nov 6 '18 at 21:12
  • $\begingroup$ @K.M I'm not sure what your objection is. $\endgroup$ – egreg Nov 6 '18 at 22:35
  • 1
    $\begingroup$ I'm not objecting. I'm simply pointing out that some of the information presented in your explanation may actually not be needed. $\endgroup$ – K.M Nov 6 '18 at 22:59



Use the product rule, $\frac{d}{dx}\left(uv\right)=v\frac{du}{dx}+u\frac{dv}{dx}$, where $u=x^2$ and $v=\sin\left(\frac{1}{x}\right)$:


Using the chain rule, $\frac{d}{dx}(\sin(\frac{1}{x}))=\frac{d\sin\left(u\right)}{du}\frac{du}{dx}$, where $u=\frac{1}{x}$ and $\frac{d}{du}(\sin(u))=\cos(u)$:


Use the power rule, $\frac{d}{dx}\left(x^n\right)=nx^{n-1}$, where $n=-1$: $\frac{d}{dx}\left(\frac{1}{x}\right)=\frac{d}{dx}\left(x^{-1}\right)=-x^{-2}$:


Simplify the expression:


Use the power rule, $\frac{d}{dx}\left(x^n\right)=nx^{n-1}$, where $n=2$: $\frac{d}{dx}\left(x^2\right)=2x$:

Answer: $$=-\cos\left(\frac{1}{x}\right)+\boxed{2x}\sin\left(\frac{1}{x}\right)$$

  • $\begingroup$ The answer doesn't really use the derivative definition, but it's still a thorough explanation of the product and chain rule $\endgroup$ – K.M Nov 4 '18 at 19:17

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