Question

What is the derivative of this function `y=e^x(sec^-1x)`?

Answer 1

`y'=e^x(sec^-1x+\frac{1}{\abs{x}\sqrt{x^2-1}})`

Explanation:

The differentiation rule for `sec^-1(x)` is `\frac{d}{dx}[sec^-1(x)]=\frac{1}{\abs{x}\sqrt{x^2-1}}`.

Using this, we can differentiate:
`y=e^x(sec^-1x)`
`y'=[e^x]'\cdot(sec^-1x) + e^x\cdot[sec^-1x]'`
`y'=e^x(sec^-1x)+e^x(\frac{1}{\abs{x}\sqrt{x^2-1}})`
`y'=e^x(sec^-1x+\frac{1}{\abs{x}\sqrt{x^2-1}})`