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Question Number 217664 by hardmath last updated on 17/Mar/25

$$\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{2\mathrm{x}+3}$$$$\mathrm{prove}\mathrm{that}:\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{1}{\sqrt{2\mathrm{x}+3}}$$

Answered by SdC355 last updated on 18/Mar/25

$$f^{\left(1\right)}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}.$$$$\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{2x+3}=\underset{h\to 0}{\lim }\frac{\sqrt{2(x+h)+3}-\sqrt{2x+3}}{h}$$$$\underset{h\to 0}{\lim }\frac{\sqrt{2x+2h+3}-\sqrt{2x+3}}{h}=$$$$\underset{h\to 0}{\lim }\frac{(\sqrt{2x+2h+3}-\sqrt{2x+3})}{h}\cdot \frac{(\sqrt{2x+2h+3}+\sqrt{2x+3})}{(\sqrt{2x+2h+3}+\sqrt{2x+3})}$$$$\therefore \underset{h\to 0}{\lim }\frac{2x+2h+3-(2x+3)}{h(\sqrt{2x+2h+3}+\sqrt{2x+3})}$$$$\underset{h\to 0}{\lim }\frac{\cancel{2x}+2h+\cancel{3}-\cancel{2x}-\cancel{3}}{h(\sqrt{2x+2h+3}+\sqrt{2x+3})}=$$$$\underset{h\to 0}{\lim }\frac{2\cancel{h}}{\cancel{h}(\sqrt{2x+h+3}+\sqrt{2x+3}}=\frac{\cancel{2}}{\cancel{2}\sqrt{2x+3}}$$$$\frac{1}{\sqrt{2x+3}}$$

Commented by hardmath last updated on 21/Mar/25

$$\mathrm{thankyou}\mathrm{dearSir}$$

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