How do you differentiate #f(x)=cot(1/sqrt(x-3)) # using the chain rule?

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Answer 1 · 2015-12-24T17:31:26

Applying a "two-degrees" chain rule.

Just remembering the concept: #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#

Here, we can name #u=(v)^(-1/2)# and #v=x-3#

Also, we must remember that the rule to derivate #cota# is

#cota=-a'csc^2a#

#(dy)/(dx)=-u'csc^2u*(-1/(2v^(3/2)))*(1)#

#(dy)/(dx)=-(-1/(2v^(3/2)))csc^2(v^(-1/2))*(-1/(2(x-3)^(3/2)))#

#(dy)/(dx)=(csc^2(x-3)^(-1/2))/(2(x-3)^(3/2))(-1/(2(x-3)^(3/2)))#

#(dy)/(dx)=-(csc^2(1/(x-3)^(1/2)))/(4(x-3)^3)#

#(dy)/(dx)=-(csc^2(1/(sqrt(x-3))))/(4(x-3)^3)#