Question

How do you differentiate `f(x)=sqrtcos(1/(2x)^3)` using the chain rule?

Answer 1

`(dy)/dx=3/16sqrt(sec(1/8x^3)tan(1/8x^3))`

Explanation:

Given:
`f(x)=sqrt(cos(1/(2x)^3)`

let `y=f(x)`

`y=sqrt(cos(1/(2x)^3)`

`y=sqrts`

`(dy)/dx=1/(2sqrts)(ds)/dx`

where,

`s=cos(1/(2x)^3)`

`s=cost`

`(ds)/dx=-sint(dt)/dx`

where,

`t=1/(2x)^3`

`t=1/u`

`(dt)/dx=-1/u^2(du)/dx`

where,

`u=(2x)^3`

`u=v^3`

`(du)/dx=3v^2(dv)/dx`

where,

`v=2x`

`(dv)/dx=2`

`(du)/dx=3v^2(dv)/dx`

`(du)/dx=3(2x)^2(2)`

`(du)/dx=24x^2`

`(dt)/dx=-1/u^2(du)/dx`

`(dt)/dx=-1/((2x)^3)^2(24x^2)`

`(dt)/dx=-1/(64x^6)(24x^2)`

`(dt)/dx=-3/8x^4`

`(ds)/dx=-sint(dt)/dx`

`(ds)/dx=-sin(1/(2x)^3)(-3/8x^4)`

`(ds)/dx=3/8sin(1/8x^3)`

`(dy)/dx=1/(2sqrt(cos(1/(2x)^3)))(3/8sin(1/8x^3))`

`(dy)/dx=3/16(sin(1/8x^3))/sqrt(cos(1/8x^3))`

`(dy)/dx=3/16sqrt((sin^2(1/8x^3))/(cos(1/8x^3)))`

`(dy)/dx=3/16sqrt(sec(1/8x^3)tan(1/8x^3))`