How do I find the derivative of $$\sqrt{x}+\frac{1}{\sqrt{x}}$$ using the first principle.
Please help me, how to go step by step. Thanks
What is the derivative of $\sqrt{x}+\frac{1}{\sqrt{x}}$ using the first principle?
0
$\begingroup$
calculus
derivatives
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4Which first principle? – 2017-02-22
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0rewriting to $x^{1/2}+x^{-1/2}$ might already do the trick for you – 2017-02-22
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0@Glorfindel **THE** first principle... :-) – 2017-02-22
2 Answers
8
Let $f(x)=\sqrt{x}+\dfrac{1}{\sqrt{x}}$
$$f^{\prime}(x) = \displaystyle\lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} $$
$$= \displaystyle\lim_{h \rightarrow 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h} + \displaystyle\lim_{h \rightarrow 0} \dfrac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h} $$
$$= \displaystyle\lim_{h \rightarrow 0} \dfrac{1}{\sqrt{x+h}+\sqrt{x}} - \displaystyle\lim_{h \rightarrow 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h\sqrt{x+h}\sqrt{x}} $$
$$= \dfrac{1}{2\sqrt{x}} - \displaystyle\lim_{h \rightarrow 0} \dfrac{1}{\sqrt{x+h}\sqrt{x}\big(\sqrt{x+h}+\sqrt{x}\big)} $$
$$= \dfrac{1}{2\sqrt{x}} - \dfrac{1}{2x\sqrt{x}}$$
$$= \dfrac{x-1}{2x\sqrt{x}}$$
1
Hint:
I'am doing the computation at $x=1$.
From $$\sqrt t-\sqrt1+\frac1{\sqrt t}-\frac1{\sqrt1}=\frac{t-1}{\sqrt t+\sqrt1}+\frac{1-t}{\sqrt1\sqrt t(\sqrt1+\sqrt t)},$$
We deduce
$$\lim_{t\to1}\frac{f(t)-f(1)}{t-1}=\frac1{\sqrt1+\sqrt1}-\frac1{\sqrt1\sqrt1(\sqrt1+\sqrt1)}.$$
Up to you to generalize for other $x$.