How do you use the chain rule to differentiate #y=sqrt(-x^4-1)(-x-2)#?

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Answer 1 · 2016-11-01T20:31:11

#y'=(3x^4+4x^3+1)/sqrt(-x^4-1)#

#y=sqrt(-x^4-1)*(-x-2)=(-x^4-1)^(1/2)(-x-2)#

#f=(-x^4-1)^(1/2), g=-x-2#

#f'=1/2(-x^4-1)^(-1/2)*-4x^3, g'=-1#

#y'=fg'+gf'#

#y'=-sqrt(-x^4-1)+(2x^4+4x^3)/sqrt(-x^4-1)#

#y'=(-(-x^4-1)+2x^4+4x^3)/sqrt(-x^4-1)#

#y'=(x^4+1+2x^4+4x^3)/sqrt(-x^4-1)#

#y'=(3x^4+4x^3+1)/sqrt(-x^4-1)#