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

1 Answer
smartspot2
Jun 30, 2017

#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}})#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
4409 views around the world
You can reuse this answer
Creative Commons License