Question

If `f(x) =sec^3(x/2)` and `g(x) = sqrt(2x-1`, what is `f'(g(x))`?

Answer 1

`f'(g(x))=3/2sec^3(sqrt(2x-1)/2)tan(sqrt(2x-1)/2)`

Explanation:

First, we should just find `f'(x)` so that we can then plug `g(x)` into it.

To find `f'(x)`, the primary and overriding issue is the third power. According to the chain rule, `d/dx[u^3]=3u^2*u'`, hence

`f'(x)=3sec^2(x/2)*d/dx[sec(x/2)]`

Now, to differentiate the secant function, recall that (through the chain rule) `d/dx[sec(u)]=sec(u)tan(u)*u'`, hence

`f'(x)=3sec^2(x/2)*sec(x/2)tan(x/2)*d/dx[x/2]`

Note that `d/dx[x/2]=d/dx[1/2x]=1/2`. Also, recognize that the secant functions can be multiplied with one another, yielding a simplified `f'(x)`:

`f'(x)=3/2sec^3(x/2)tan(x/2)`

To find `f'(g(x))`, simply plug in `sqrt(2x-1)` wherever there's an `x` in `f'(x)`.

`f'(g(x))=3/2sec^3(sqrt(2x-1)/2)tan(sqrt(2x-1)/2)`

Answer 2

`f'(g(x))=3/(2(sqrt(2x-1)))sec^3(sqrt(2x-1)/2)*tan(sqrt(2x-1)/2)`

Explanation:

If the question meaning is `f'(g(x))` as `(f@g)'` then:

find `f@g=f(g(x))`

`y=f(g(x))=[sec(sqrt(2x-1)/2)]^3`

`g(x)` is a composition too, therefore you can think:

`y=u^3`

`u=sec(v)`

`v=sqrt(z)/2`

`z=(2x-1)`

the Chain Rule tells

`dy/dx=f'(g(x))=(dy)/(du)(du)/(dv)(dv)/(dz)(dz)/(dx)`

`dy/(du)=3u^2=3sec^2(v)=3sec^2(sqrt(z)/2)=3sec^2(sqrt(2x-1)/2)`

`(du)/(dv)=sec'(v)=tan(v)*sec(v)=tan(sqrt(z)/2)*sec(sqrt(z)/2)=`
`=tan(sqrt(2x-1)/2)*sec(sqrt(2x-1)/2)`

`(dv)/(dz)=1/2*(1/(2sqrt(z)))=1/(4sqrt(2x-1))`

`(dz)/(dx)=2*1+0`

Thus:

`y'=3sec^2(sqrt(2x-1)/2)*tan(sqrt(2x-1)/2)*sec(sqrt(2x-1)/2)*1/(cancel(4)^2sqrt(2x-1))*cancel(2)=`
`=3/(2(sqrt(2x-1)))sec^2(sqrt(2x-1)/2)*tan(sqrt(2x-1)/2)*sec(sqrt(2x-1)/2)`

Alternatively, once you get practice, more simply:

Given:

`h(x)=f(g(x))`

`h'(x)=f'(g(x))*g'(x)`

if `g(x)=g(i(x)) =>g'(x)=g'(i(x))*i'(x)`

thus you can apply the rule ricorsively.