Find dy/dx where y=1/4v^4 and v=2/3x^3+5?

Question

1647 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-26T17:06:59

#dy/dx=2x^2(2/3x^3+5)^3#

Hope the question is:

#y=1/4v^4#

#v=2/3x^3+5#

#dy/dx=dy/(dv)(dv)/dx#

Here

#dy/(dv)=1/4xx4v^3=v^3#
#dy/(dv)=(2/3x^3+5)^3#
#(dv)/(dx)=2/3xx3x^2+0=2x^2#

Thus,

#dy/dx=(2/3x^3+5)^3(2x^2)#

#dy/dx=2x^2(2/3x^3+5)^3#

Answer 2 · 2018-02-26T17:07:06

#(dy)/(dx)=2x^2(2/3x^3+5)^3#

As #y=1/4v^4#, #(dy)/(dv)=1/4xx4v^3=v^3#

and as #v=2/3x^3+5#, #(dv)/(dx)=2/3xx3x^2=2x^2#

Hence #(dy)/(dx)=(dy)/(dv)xx(dv)/(dx)#

= #v^3xx2x^2#

= #(2/3x^3+5)^3xx2x^2#

= #2x^2(2/3x^3+5)^3#