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Question Number 95339 by mathocean1 last updated on 24/May/20

$$\mathrm{f}\left(\mathrm{x}\right)=\mid 2\mathrm{x}+3\mid$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=...?$$

Answered by john santu last updated on 24/May/20

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{(2\mathrm{x}+3)}^2}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{2{(2\mathrm{x}+3)}^1.2}{2\sqrt{{(2\mathrm{x}+3)}^2}}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{4\mathrm{x}+6}{\mid 2\mathrm{x}+3\mid }$$

Answered by prakash jain last updated on 24/May/20

$$x=\frac{y}{2}-\frac{3}{2}\Rightarrow y=2(x+\frac{3}{2})$$$$\mid 2x-3\mid =\mid y\mid$$$$\frac{df\left(x\right)}{dx}=\frac{d}{dy}\mid y\mid \times \frac{dy}{dx}=\frac{y}{\mid y\mid }\times 2$$$$=\frac{2(x+\frac{3}{2})}{\mid 2(x+\frac{3}{2})\mid }\times 2=\frac{4x+6}{\mid 2x+3\mid }$$

Commented by john santu last updated on 24/May/20

$$\mathrm{yes}.\mathrm{your}\mathrm{are}\mathrm{right}$$

Answered by  M±th+et+s last updated on 24/May/20

$$byfirstprinciple$$$$\underset{h\to 0}{lim}\frac{\mid 2(x+h)+3\mid -\mid 2x+3\mid }{h}.\frac{\mid 2(x+h)+3\mid +\mid 2x+3\mid }{\mid 2(x+h)+3\mid +\mid 2x+3\mid }$$$$\underset{h\to 0}{lim}\frac{\mid (2x+3)+2h{\mid }^2-\mid 2x+3{\mid }^2}{h(\mid 2(x+h)+3\mid +\mid 2x+3\mid )}$$$$\underset{h\to 0}{lim}\frac{{(2x+3)}^2+4(2x+3)h+4h^2-{(2x+3)}^2}{h(\mid (2x+3)+2h\mid +\mid 2x+3\mid )}$$$$\underset{h\to 0}{lim}\frac{4h(2x+3+h)}{h(\mid (2x+3)+2h\mid +\mid 2x+3\mid )}$$$$\underset{x\to 0}{lim}\frac{4(2x+3+h)}{\mid 2x+3+h\mid +\mid 2x+3\mid }=\frac{4x+6}{\mid 2x+3\mid }$$$$$$

Commented by john santu last updated on 24/May/20

$$\mid 2(\mathrm{x}+\mathrm{h})+3\mid =\mid (2\mathrm{x}+3)+2\mathrm{h}\mid$$

Commented by  M±th+et+s last updated on 24/May/20

$$yesi{didn}^{\prime }tfocus.thankyousir$$

Answered by mathmax by abdo last updated on 24/May/20

$$\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}2\mathrm{x}+3\mathrm{if}\mathrm{x}>-\frac{3}{2}\mathrm{and}\mathrm{f}(-\frac{3}{2})=0\Rightarrow \\ -2\mathrm{x}-3\mathrm{if}\mathrm{x}<-\frac{3}{2}\end{array}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\{\begin{array}{l}2\mathrm{if}\mathrm{x}>-\frac{3}{2}\\ -2\mathrm{if}\mathrm{x}<-\frac{3}{2}\end{array}$$$$\mathrm{rest}\mathrm{studying}\mathrm{derivability}\mathrm{at}{\mathrm{x}}_0=-\frac{3}{2}....$$

Answered by 1549442205 last updated on 25/May/20

$$\mathrm{we}\mathrm{have}{\mathrm{f}}^2\left(\mathrm{x}\right)={4\mathrm{x}}^2+12\mathrm{x}+9\Rightarrow$$$$2\mathrm{f}\left(\mathrm{x}\right).\mathrm{f}{}^{{}^{\prime }}\left(\mathrm{x}\right)=8\mathrm{x}+12\Rightarrow \mathrm{f}{}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{8\mathrm{x}+12}{2\mathrm{f}\left(\mathrm{x}\right)}=\frac{4\mathrm{x}+6}{\mid 2\mathrm{x}+3\mid }$$

Answered by mathmax by abdo last updated on 10/Oct/21

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{(2\mathrm{x}+3)}^2}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{2.2(2\mathrm{x}+3)}{2\sqrt{{(2\mathrm{x}+3)}^2}}=\frac{4\mathrm{x}+6}{\mid 2\mathrm{x}+3\mid }$$

Commented by prakash jain last updated on 10/Oct/21

$$\mathrm{how}\mathrm{are}\mathrm{you},\mathrm{sir}$$

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