How do you find the derivative of #s=[(t^2+1)^3+t]^-1#?

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Answer 1 · 2017-03-14T01:59:20

#(ds)/dt =- (6t(t^2+1)^2+1)/((t^2+1)^3+t)^2#

#s=[(t^2+1)^3+t]^-1#

#=1/((t^2+1)^3+t)#

#(ds)/dt =(0-1*d/dt((t^2+1)^3+t))/((t^2+1)^3+t)^2#

Applying the chain rule

#=-(3(t^2+1)^2*2t+1)/((t^2+1)^3+t)^2#

#=- (6t(t^2+1)^2+1)/((t^2+1)^3+t)^2#