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So the function is $f(x) = \sqrt{x+2} -x$, and I keep hitting dead ends trying to solve it using the definition of derivative. If anyone can help, it would be greatly appreciated!

Hadi
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ahhh
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  • How far do you get? Where do you get stuck? – Arthur Mar 17 '20 at 17:40
  • I manage to input everything into the definition, but the confusion hits after I rationalise the numerator? I can't cancel out my denominator :( – ahhh Mar 17 '20 at 17:44

2 Answers2

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$\require{cancel}$Note that $$f(x)=\sqrt{x+2}-x$$ $$f(x+h)=\sqrt{x+h+2}-x-h$$ $$f(x+h)-f(x)=\sqrt{x+h+2}-\sqrt{x+2}-h$$ So we have $$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}-h}{h}$$ $$= \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}-1$$ $$= \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}\cdot \frac{\sqrt{x+h+2} + \sqrt{x+2}} {\sqrt{x+h+2}+\sqrt{x+2}} - 1$$ $$=\frac{(\cancel{x}+h+\cancel{2})-(\cancel{x}+\cancel{2})}{h(\sqrt{x+h+2} + \sqrt{x+2})} - 1$$ $$=\frac{\cancel{h}}{\cancel{h}(\sqrt{x+h+2} + \sqrt{x+2})} - 1$$ $$= \frac{1}{\sqrt{x+h+2} + \sqrt{x+2}} - 1$$ Let $h\to 0$ to get $$\boxed{f’(x) = \frac{1}{2\sqrt{x+2}}-1}$$

MPW
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Let’s look at

$$\begin{align} \Delta(x,h)&={\sqrt{x+h+2}-\sqrt{x+2}-h\over h}\\ &={\sqrt{x+h+2}-\sqrt{x+2}\over h}-{h\over h}\\ &={h\over h\left(\sqrt{x+h+2}+\sqrt{x+2}\right)}-1 \end{align}$$

Where the last equality is the result of multiplying numerator and denominator by the conjugate of $\sqrt{x+h+2}-\sqrt{x+2}$.

And we can now get the limit

$$\lim_{h\to 0}\Delta(x,h)={1\over 2\sqrt{x+2}}-1$$

marwalix
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