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

1 Answer
Dec 30, 2015

Answer:

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

Explanation:

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)#