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Question Number 118887 by rexfordattacudjoe last updated on 20/Oct/20

Answered by benjo_mathlover last updated on 20/Oct/20

$$f{}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}$$$$=\underset{h\to 0}{\lim }\frac{x+h-x}{h}\times \underset{h\to 0}{\lim }\frac{1}{\sqrt{x+h}+\sqrt{x}}$$$$=1\times \frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}}$$

Answered by ebi last updated on 20/Oct/20

$$f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{\mathrm{li}m}\frac{f(x+h)-f\left(x\right)}{h}$$$$f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{\mathrm{li}m}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$$$f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{\mathrm{li}m}\frac{\sqrt{x+h}-\sqrt{x}}{h}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$$$$f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{\mathrm{li}m}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$$$$f^{{}^{\prime }}\left(x\right)=\frac{1}{\sqrt{x+0}+\sqrt{x}}$$$$f^{{}^{\prime }}\left(x\right)=\frac{1}{2\sqrt{x}}$$

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