How do you find the derivative of the function: #arcsec(x/2)#?

1 Answer
sente
May 2, 2016

#d/dx"arcsec"(x/2)=2/(xsqrt(x^2-4))#

Explanation:

Using implicit differentiation, we start by letting #y = "arcsec"(x/2)#

#=> sec(y) = x/2#

#=> d/dxsec(y) = d/dxx/2#

#=>sec(y)tan(y)dy/dx = 1/2#

#=> dy/dx = 1/(2sec(y)tan(y))#

We already know #sec(y) = x/2#, and if we construct a right triangle with an angle #y# such that #sec(y) = x/2# we find that #tan(y) = sqrt(x^2-4)/2#. Substituting these in, we have

#d/dx"arcsec"(x/2) = dy/dx#

#= 1/(2(x/2)(sqrt(x^2-4)/2))#

#=2/(xsqrt(x^2-4))#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
4581 views around the world
You can reuse this answer
Creative Commons License