if-y-x-2-x-1-find-dy-dx-from-first-principle-

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

i f y = \frac{x + 2}{\sqrt{x + 1}} f i n d \frac{d y}{d x} f r o m f i r s t

p r i n c i p l e .

Answered by bemath last updated on 14/Oct/20

y = \frac{x + 1 + 1}{\sqrt{x + 1}} = \sqrt{x + 1} + \frac{1}{\sqrt{x + 1}}

\frac{dy}{dx} = \lim_{h \to 0} \frac{\sqrt{x + h + 1} + \frac{1}{\sqrt{x + h + 1}} - \sqrt{x + 1} - \frac{1}{\sqrt{x + 1}}}{h}

= \lim_{h \to 0} \frac{h}{h (\sqrt{x + h + 1} + \sqrt{x + 1})} + \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h + 1}} - \frac{1}{\sqrt{x + 1}}}{h}

= \frac{1}{2 \sqrt{x + 1}} + \lim_{h \to 0} \frac{\sqrt{x + 1} - \sqrt{x + h + 1}}{h (\sqrt{(x + 1) (x + h + 1)}}

= \frac{1}{2 \sqrt{x + 1}} + \frac{1}{x + 1} . \lim_{x \to 0} \frac{- h}{h (\sqrt{x + h + 1} + \sqrt{x + 1})}

= \frac{1}{2 \sqrt{x + 1}} - \frac{1}{x + 1} . \frac{1}{2 \sqrt{x + 1}}

= \frac{1}{2 \sqrt{x + 1}} [1 - \frac{1}{x + 1}] = \frac{x}{2 {(x + 1)}^{\frac{3}{2}}}

or \frac{x}{2 (x + 1) \sqrt{x + 1}}

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