How do you differentiate # y =1/sqrtln(x^2-3x)# using the chain rule?

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Answer 1 · 2015-12-31T12:42:53

We can rewrite it as #y=(ln(x^2-3x))^(-1/2)# and use the chain rule, which states that #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#

Renaming:

#y=u^(-1/2)#
#u=ln(v)#
#v=x^2-3x#

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

#(dy)/(dx)=-(2x-3)/(2u^(3/2)v)#

Substituting #u#:

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

Substituting #v#:

#(dy)/(dx)=-(2x-3)/(2(ln(x^2-3x))^(3/2)*(x^2-3x))#

Finally:

#(dy)/(dx)=(3-2x)/(2(x^2-3x)sqrt((ln(x^2-3x))^3#