Question

How do you differentuate `y^2=sin^4(2x)+ cos^4(2x)`?

Answer 1

`(dy)/(dx)=(4sin^3(2x)cos(2x))/y-(4cos^3(2x)sin(2x))/y`

Explanation:

By using implicit differentiation and chain rule

`y^2=sin^4(2x)+cos^4(2x)`

`2ytimes(dy)/(dx)=4sin^3(2x)times2cos(2x)-4cos^3(2x)times2sin(2x)`

`2ytimes(dy)/(dx)=8sin^3(2x)cos(2x)-8cos^3(2x)sin(2x)`

`(dy)/(dx)=(4sin^3(2x)cos(2x))/y-(4cos^3(2x)sin(2x))/y`

For the chain rule, just to expand on what I did:

let `z=sin^4(2x)`
and let `u=2x`
so `(du)/(dx)=2`

Okay, so `z=sin^4u`
`(dz)/(du)=4sin^3utimescosu=4sin^3ucosu`

Now `(dz)/(dx)=(dz)/(du)times(du)/(dx)`
So, `(dz)/(dx)=4sin^3ucosutimes2`

`(dz)/(dx)=8sin^3ucosu`
`(dz)/(dx)=8sin^3(2x)cos(2x)`

I applied the same format for `cos^4(2x)`