How to derive $y=\sqrt{x+\sqrt{x^2+5}}$ at $x=2$.I used logarithmic differentiation and chain rule over and over again but I can't get the right answer
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\begin{align*} \frac{d}{dx}y & = \frac{d}{dx}\sqrt{x+\sqrt{x^2+5}}\\ & = \frac{1}{2}\frac{1}{\sqrt{x+\sqrt{x^2+5}}}\cdot \frac{d}{dx}(x+\sqrt{x^2+5})\\ & = \frac{1}{2}\frac{1}{\sqrt{x+\sqrt{x^2+5}}} \,\left(1+\frac{x}{\sqrt{x^2+5}}\right)\\ \end{align*} Now simplify as best as you can.
Anurag A
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Ahaaaa! I'm grateful . – user162143 Jul 07 '14 at 17:38
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There is a name to this rule, I keep forgetting it. Is it called the chain rule? – MonK Jul 07 '14 at 17:44
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$$f(x)=\sqrt{x+\sqrt{x^2+5}}$$
Then $$f'(x)=\frac{1+\frac{2x}{2\sqrt{x^2+5}}}{2\sqrt{x+\sqrt{x^2+5}}}=$$
tattwamasi amrutam
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1This gives the answer, but it looks like the difficulty that the OP is having is with the process, which isn't cleared up at all by just giving the final answer. – qaphla Jul 07 '14 at 17:17
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Another useful way to do derivatives like these is implicitly, using $$\frac{d}{dx}y^2=\frac{d}{dx}\left(x+\sqrt{x^2+5}\right)$$ This is particularly useful when it's something more complicated than a simple square root, but a couple simple steps will get you to the correct answer. $$2y\frac{dy}{dx}=1+\frac{1}{2}\cdot\frac{2x}{\sqrt{x^2+5}}\\\frac{dy}{dx}=\left(1+\frac{x}{\sqrt{x^2+5}}\right)\frac{1}{2\sqrt{x+\sqrt{x^2+5}}}$$
Silynn
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