What will be the derivative of x[x] when x is not an integer?
I was applying the product rule but I'm stuck in the part were I have to differentiate [x] wrt x.
[.] Denotes greatest integer function
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What does the notation "[.]" denote? – Bobson Dugnutt Mar 09 '17 at 11:59
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Greatest integer function – user423865 Mar 09 '17 at 12:00
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3What does the graph of $[x]$ look like, when $x$ is not an integer. Doesn't that tell you what the derivative is? – Gerry Myerson Mar 09 '17 at 12:00
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Ok I get it that's 0 – user423865 Mar 09 '17 at 12:04
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@GerryMyerson thanks – user423865 Mar 09 '17 at 12:04
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Hint: on intervals of the form $(n,n+1)$ where $n$ is an integer the function $x\mapsto x\lfloor x\rfloor$ coincides with $x\mapsto nx$ (which is easy to differentiate). I would leave out the product rule. – drhab Mar 09 '17 at 12:08
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1Just to be clear, the answer to your question is not $0$. Suppose $2<x<3$, just to pick. Then $\lfloor x \rfloor=2$ so in that interval $f(x)=2x$. The derivative of $\lfloor x \rfloor$ is $0$ (where it is defined). – lulu Mar 09 '17 at 12:08
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The derivative is defined in $\Bbb R\setminus\Bbb Z$. – Masacroso Mar 09 '17 at 12:21
2 Answers
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You must evaluate: $$\frac{d}{dx} ([x]\cdot x) \tag{1}$$ Your idea to use the product rule is a good one. Therefore: $$\frac{d}{dx} ([x]\cdot x)=[x]\cdot \frac{d}{dx} (x)+x\cdot \frac{d}{dx}([x]) \tag{2}$$ Note that $[x]$ is locally constant for $x\not\in \mathbb{Z}$. It follows that $\frac{d}{dx}([x])=0$ where $x\not\in \mathbb{Z}$.
Therefore, what can you deduce about the solution to $(1)$ using $(2)$?
projectilemotion
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$$\frac{d}{dx} (x[x]) = [x]\cdot \frac{d}{dx} (x) + x\cdot \frac{d}{dx} ([x]) =[x]\cdot 1 + x\cdot 0 =[x] $$
sgrmshrsm7
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