What is the derivative: #g(y) = sqrt[2y + (3y + 4y^2)^3]#?

Question

I got to
#[2 + (9 + 24y)(3y + 4y^2)^2]/{2 sqrt [2y + (3y + 4y^2)^3]}#.

1880 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-07T13:48:38

Your answer is right

We are given #g(y)=sqrt[2y+(3y+4y^2)^3]=(2y+(3y+4y^2)^3)^(1/2)#

#g(y)=d/dy[(2y+(3y+4y^2)^3)^(1/2)]#
#=1/2*(2y+(3y+4y^2)^3)^(1/2-1)* d/dy[2y+(3y+4y^2)^3]#

#=(2y+(3y+4y^2)^3)^(-1/2)/2* (d/dy[2y]+d/dy[(3y+4y^2)^3])#

#=(2y+(3y+4y^2)^3)^(-1/2)/2* (2+(3*(3y+4y^2)^(3-1)d/dy[3y+4y^2]))#

#=(2y+(3y+4y^2)^3)^(-1/2)/2* (2+3(3y+4y^2)^2(3+8y))#

#=((2y+(3y+4y^2)^3)^(-1/2)(2+3(3y+4y^2)^2(3+8y)))/2#

#=(2+3(3y+4y^2)^2(3+8y))/(2(2y+(3y+4y^2)^3)^(1/2))#

#=(2+(9+24y)(3y+4y^2)^2)/(2sqrt(2y+(3y+4y^2)^3))#