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)=\frac1{\sec f(x)\tan f(x)}`

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


`(\sec^(-1))'(\sqrt2)=f'(\sqrt2)=\frac1{\sec\left(\sec^(-1)\sqrt2\right)\tan\left(\sec^(-1)\sqrt2\right)}`

You have
`\sec^(-1)\sqrt2=\frac\pi4`, so
`\sec\left(\sec^(-1)\sqrt2\right)=\sqrt2` and
`\tan\left(\sec^(-1)\sqrt2\right)=1`. So
`(\sec^(-1))'(\sqrt2)=\frac1{\sqrt2*1}=\frac1{\sqrt2}`