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$\DeclareMathOperator{\arcsec}{arcsec}$How is $\arcsec x$ differentiated using first principles? I tried substituting $A = \arcsec(x+h)$ and $B = \arcsec(x)$ while solving using $h \to 0$, but it doesn't seem to be a good idea (as $\arcsec(\sec x)$ is not always equal to $x$).

Could anyone please post a detailed solution?

P.S. The original question asked for the derivative of $\sqrt{\arcsec x}$; I managed to bring it till here after multiplying the denominator and numerator by conjugate of numerator and eliminating the square root. Help is appreciated. Thanks a lot.

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    $\begingroup$ Why not use the inverse function derivative formula? $\endgroup$ – Ofek Gillon May 2 '17 at 11:33
  • $\begingroup$ Your concern about $\arcsec(\sec(x))$ can be obviated; if you restrict $x$ to the range of $\arcsec(u)$, then $\arcsec(\sec(x)) = x$ is always true. $\endgroup$ – user14972 May 2 '17 at 14:12
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Let $\text{arcsec}(x+h)=2A,\text{arcsec}(x)=2B\implies\sec2B=x,\cos2B=?$

Using principal values, $\sin2B=+\dfrac1{\sqrt{1-\dfrac1{x^2}}}=+\dfrac x{\sqrt{x^2-1}}$

$$\lim_{h\to0}\dfrac{\text{arcsec}(x+h)-\text{arcsec}(x)}h=\lim_{A-B\to0}\dfrac{2A-2B}{\sec2A-\sec2B}$$

$$=\lim_{A-B\to0}\dfrac{(2A-2B)\cos2A\cos2B}{\cos2B-\cos2A} =\lim_{A-B\to0}\dfrac{(2A-2B)\cos2A\cos2B}{2\sin(A-B)\sin(A+B)}=\dfrac{\cos^22B}{\sin2B}=?$$

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Hint: you can use the relation$\DeclareMathOperator{\arcsec}{arcsec}$ $$\arcsec x=\frac{\pi} {2}-\arcsin\left(\frac{1}{x}\right)$$ and the formula $$\arcsin a - \arcsin b=\arcsin(a\sqrt{1-b^{2}}-b\sqrt{1-a^{2}})$$ Finally you will also need the standard limit $$\lim_{x\to 0}\frac{\arcsin x} {x} =1$$

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