How do you differentiate #f(x)=cot(sqrt(x^2-1)) # using the chain rule?

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Answer 1 · 2015-12-22T06:21:04

The explanation is given below.

#f(x) =cot(sqrt(x^2-1))#

Let us first write the same as the following.

#y = cot(u)# and #u=sqrt(v)# and #v=x^2-1#
The above steps are like breaking the chain into manageable links.

Chain rule would work like this:

#dy/dx = dy/(du) * (du)/(dv) * (dv)/(dx)# Chain rule

#y=cot(u)#
#dy/(du) = -sec^2(u)#

#u=sqrt(v)#
#(du)/(dv) = 1/(2sqrt(v))#

#v=x^2-1#
#(dv)/(dx) = 2x#

# f'(x) = dy/dx = -sec^2(u) * 1/(2sqrt(v)) * 2x#
Simplifying

#f'(x) = dy/dx =(- x*sec^2(sqrt(v)))/sqrt(v) #
#f'(x) = dy/dx = (-x*sec^2(sqrt(x^2-1))/sqrt(x^2-1))#

#f'(x) = (-x*sec^2(sqrt(x^2-1))/sqrt(x^2-1))# answer