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Question Number 3385 by Filup last updated on 12/Dec/15

$$\mathrm{let}:$$$$S=\underset{i=a}{\sum ^n}{x}_i$$$$\mathrm{prove}\mathrm{or}\mathrm{disprove}\mathrm{that}:$$$$\frac{d}{dx}\left(\underset{i=a}{\sum ^n}{x}_i\right)=\underset{i=a}{\sum ^n}\left(\frac{d}{dx}\left\{x_i\right\}\right)$$$$=\underset{i=a}{\sum ^n}{x}_i^{{}^{\prime }}$$

Commented by prakash jain last updated on 12/Dec/15

$$\mathrm{Some}\mathrm{of}\mathrm{two}\mathrm{function}\mathrm{which}\mathrm{are}\mathrm{not}\mathrm{differentiable}$$$$\mathrm{at}x=x_0\mathrm{may}\mathrm{be}\mathrm{differentiable}\mathrm{at}x=x_0.$$

Commented by Filup last updated on 12/Dec/15

$$iwasmeanttowritethat$$$$x_{i}isthe{i}^{\mathrm{th}}terminaseries$$

Commented by prakash jain last updated on 12/Dec/15

$$\mathrm{It}\mathrm{does}\mathrm{make}\mathrm{a}\mathrm{difference}.\mathrm{If}{x}_i{i}^{th}\mathrm{term}\mathrm{in}$$$$\mathrm{series}.\mathrm{It}\mathrm{is}\mathrm{a}\mathrm{still}\mathrm{function}\mathrm{of}x.$$

Commented by Filup last updated on 12/Dec/15

$$\mathrm{i}\mathrm{see}.\mathrm{hmm}...\mathrm{so}\mathrm{is}\mathrm{there}\mathrm{ever}\mathrm{a}\mathrm{time}$$$$\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{above}\mathrm{is}\mathrm{true}?$$$$\mathrm{other}\mathrm{than}{x}_i=\mathrm{constant}$$

Commented by prakash jain last updated on 12/Dec/15

$$\mathrm{It}\mathrm{is}\mathrm{generally}\mathrm{true}\mathrm{if}\mathrm{each}\mathrm{of}{x}_i\mathrm{is}\mathrm{differentiable}.$$$$\mathrm{I}\mathrm{gave}\mathrm{an}\mathrm{exception}\mathrm{rather}\mathrm{than}\mathrm{rule}.$$

Commented by Filup last updated on 12/Dec/15

$$\mathrm{Ah}\mathrm{I}\mathrm{understand}\mathrm{now},\mathrm{thanks}!$$

Commented by prakash jain last updated on 12/Dec/15

$$\frac{d}{dx}(f\left(x\right)+g\left(x\right))=\frac{d}{dx}f\left(x\right)+\frac{d}{dx}g\left(x\right)$$$$whatyouhaveaboveisaessentiallysame$$$$formula.$$

Answered by prakash jain last updated on 12/Dec/15

$$x_1=\{\begin{array}{ll}2& x⩾1\\ 3& x<1\end{array}$$$$x_2=\{\begin{array}{ll}-2& x⩾1\\ -3& x<1\end{array}$$$$\frac{d}{dx}(x_1+x_2)\mathrm{at}x=1\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}\frac{d}{dx}{x}_1+\frac{d}{dx}{x}_2$$$$\mathrm{at}x=1.$$

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