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

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Answer 1 · 2018-05-14T16:12:14

#dy/dx=[-(2x+3)*(x^2+3x)^(-3/2)]/2#

show below

#y=(x^2+3x)^(-1/2)#

let suppose:

#u=x^2+3x#

#(du)/dx=2x+3#

#y=u^(-1/2)#

#dy/(du)=-1/2*u^(-3/2)#

#dy/dx=(du)/dx*dy/(du)#

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

#dy/dx=[-(2x+3)*u^(-3/2)]/2#

#dy/dx=[-(2x+3)*(x^2+3x)^(-3/2)]/2#