How do you use the chain rule to differentiate #y=root3(2x^2+6x)#?

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Answer 1 · 2016-11-23T13:49:50

The answer is #=1/3(2x^2+6x)^(-2/3)*(4x+6)#

#y=(2x^2+6x)^(1/3)#

We use #(x^n)'=nx^(n-1)#

#(u^(1/3))'=1/3u^(-2/3)#

#(v^2)'=2v#

We use the chain rule

#dy/dx=((2x^2+6x)^(1/3))'#

#=1/3(2x^2+6x)^(-2/3)*(2x^2+6x)'#

#=1/3(2x^2+6x)^(-2/3)*(4x+6)#

Answer 2 · 2016-11-23T14:28:25

#dy/dx=1/3(2x^2+6x)^(2/3) (4x-6)#

let #u=2x^2+6x " "=>" " (du)/(dx) = 4x+6#

Let #y=u^(1/3)" "=>" " dy/(du) =1/3 u^(-2/3)#

#color(green)(dy/(du)xx(du)/(dx))" " =" " dy/dxxx(du)/(du)" " =" " color(red)(dy/dx)#

#color(white)(.)#

Thus #" "color(green)(color(red)(dy/dx=)1/3 u^(-2/3)xx(4x-6))#

but #u=2x^2+6x# so we have:

#dy/dx=1/3(2x^2+6x)^(2/3) (4x-6)#