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I'am differentiating $$y= 3\sinh^{-1}\sqrt{2x^2-1}$$

Please let me know if i've done it right.$$\frac{\mathrm{dy} }{\mathrm{d} x}=3 \left [ \frac{\frac{1}{2}(2x^2-1)^\frac{-1}{2}4x}{\sqrt{2x^2}} \right ]$$

thanks in advance.

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up vote 0 down vote accepted

Note that $\mathrm{sinh}^{-1} z = \log{(z + \sqrt{1 + z^2})}$

$$ y= 3\sinh^{-1}\sqrt{2x^2-1} = 3 \log{(z + \sqrt{1 + z^2})}$$

where $z = \sqrt{2 x^2-1}$. Then

$$\frac{dy}{dx} = \frac{3}{z + \sqrt{1+z^2}} \left ( 1 + \frac{z}{\sqrt{1 + z^2}} \right ) \frac{dz}{dx} $$

So you can verify that you are on the right track.

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For checking answers, one could consult wolfram alpha.

However, it is a very useful skill to be able to come up with says to sanity check your own work. Some ideas of how you might do that by hand are:

You could numerically estimate the derivative at several points; e.g. compute $(f(1.00001) - f(1)) / (1.00001 - 1)$ and compare to the value of the claimed derivative

You could do symbolic estimation or compute the first two terms of the Taylor series about some point by alternate means. e.g. around $x=1$ we can estimate:

$$ \sqrt{2x^2 - 1} \approx \sqrt{1 + 4(x-1)^2} \approx 1 + (1/2) \left( 4(x-1)^2 \right) $$

where the second approximation uses the Taylor series for $\sqrt{1 + t}$ about $t=0$. You can continue along until you get an estimate $y \approx a + b (x-1) $, and then compare $b$ to the value of the claimed derivative at $x=1$.

Neither of these are foolproof, but these sorts of things are very frequently useful for getting some quick sanity checks on your work.

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The online version of wolfram alpha doesn't computute all the derivatives.sometimes it doesn't even understand what i'am talking. – alok Jan 11 '13 at 5:10

$$\sqrt{2x^2-1}=\sinh\frac y3$$

$$\implies \sqrt{2x^2-1}=\sinh\frac y3\implies 2x^2=1+\sinh^2\frac y3=\cosh^2\frac y3$$

So, $$\cosh\frac y3=\sqrt2x$$

Differentiating wrt $x,$ $$\frac13 \sinh\frac y3\frac{dy}{dx}=\sqrt2$$

$$\implies \frac{dy}{dx}=\frac{3\sqrt2}{\sinh\frac y3}=\frac{3\sqrt2}{\sqrt{2x^2-1}}$$

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