How do you differentiate # y= sqrt( (x^2 + 4x + 1)^2+2x)# using the chain rule?

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Answer 1 · 2015-12-30T07:29:18

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

According to the chain rule:

#d/dx(u^(1/2))=1/2(u^(-1/2))*u'=(u')/(2sqrtu)#

Set #u=(x^2+4x+1)^2+2x#.

Using the aforementioned rule:

#y'=(d/dx((x^2+4x+1)^2+2x))/(2sqrt((x^2+4x+1)^2+2x)#

Find #d/dx((x^2+4x+1)^2+2x)#.

For the first term, use chain rule again.

#=>2(x^2+4x+1)(2x+4)+2#

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