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Question Number 117863 by aurpeyz last updated on 14/Oct/20

$$ify=\frac{x+2}{\sqrt{x+1}}find\frac{dy}{dx}fromfirst$$$$principle.$$

Answered by bemath last updated on 14/Oct/20

$$\mathrm{y}=\frac{\mathrm{x}+1+1}{\sqrt{\mathrm{x}+1}}=\sqrt{\mathrm{x}+1}+\frac{1}{\sqrt{\mathrm{x}+1}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\underset{\mathrm{h}\to 0}{\lim }\frac{\sqrt{\mathrm{x}+\mathrm{h}+1}+\frac{1}{\sqrt{\mathrm{x}+\mathrm{h}+1}}-\sqrt{\mathrm{x}+1}-\frac{1}{\sqrt{\mathrm{x}+1}}}{\mathrm{h}}$$$$=\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{h}}{\mathrm{h}(\sqrt{\mathrm{x}+\mathrm{h}+1}+\sqrt{\mathrm{x}+1})}+\underset{\mathrm{h}\to 0}{\lim }\frac{\frac{1}{\sqrt{\mathrm{x}+\mathrm{h}+1}}-\frac{1}{\sqrt{\mathrm{x}+1}}}{\mathrm{h}}$$$$=\frac{1}{2\sqrt{\mathrm{x}+1}}+\underset{\mathrm{h}\to 0}{\lim }\frac{\sqrt{\mathrm{x}+1}-\sqrt{\mathrm{x}+\mathrm{h}+1}}{\mathrm{h}(\sqrt{(\mathrm{x}+1)(\mathrm{x}+\mathrm{h}+1)}}$$$$=\frac{1}{2\sqrt{\mathrm{x}+1}}+\frac{1}{\mathrm{x}+1}.\underset{x\to 0}{\lim }\frac{-\mathrm{h}}{\mathrm{h}(\sqrt{\mathrm{x}+\mathrm{h}+1}+\sqrt{\mathrm{x}+1})}$$$$=\frac{1}{2\sqrt{\mathrm{x}+1}}-\frac{1}{\mathrm{x}+1}.\frac{1}{2\sqrt{\mathrm{x}+1}}$$$$=\frac{1}{2\sqrt{\mathrm{x}+1}}[1-\frac{1}{\mathrm{x}+1}]=\frac{\mathrm{x}}{2{(\mathrm{x}+1)}^{\frac{3}{2}}}$$$$\mathrm{or}\frac{\mathrm{x}}{2(\mathrm{x}+1)\sqrt{\mathrm{x}+1}}$$

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