Question

Help math question derivative!

Answer 1

Let $f(x)=\sec^{-1}x$. Then $\sec f(x)=x$, and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$, you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$, so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$. So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$