Question

What is the second derivative of `f(cosx)` when `x=pi/2` where `f(x)=sinx`?

Answer 1

`0`

Explanation:

We have: `f(x) = sin(x)`

`Rightarrow f(cos(x)) = sin(cos(x))`

Let's differentiate this function:

`Rightarrow f'(cos(x)) = frac(d)(dx)(sin(cos(x)))`

We can differentiate this using the chain rule.

Let `u = cos(x) Rightarrow u' = - sin(x)` and `v = sin(u) Rightarrow v' = cos(u)`:

`Rightarrow f'(cos(x)) = u' cdot v'`

`Rightarrow f'(cos(x)) = - sin(x) cdot cos(u)`

`Rightarrow f'(cos(x)) = - sin(x) cos(u)`

Replace `u` with `cos(x)`:

`Rightarrow f'(cos(x)) = - sin(x) cos(cos(x))`

We need to find the second derivative, so let's differentiate this one more time:

`Rightarrow f''(cos(x)) = frac(d)(dx)(- sin(x) cos(cos(x)))`

`Rightarrow f''(cos(x)) = - frac(d)(dx)(sin(x) cos(cos(x)))`

We can differentiate this using a combination of the product rule and the chain rule:

`Rightarrow f''(cos(x)) = sin(x) cdot frac(d)(dx)(cos(cos(x))) + cos(cos(x)) cdot frac(d)(dx)(sin(x))`

Let `u = cos(x) Rightarrow u' = - sin(x)` and `v = cos(u) Rightarrow v' = - sin(u)`:

`Rightarrow f''(cos(x)) = sin(x) cdot u' cdot v' + cos(cos(x)) cdot cos(x)`

`Rightarrow f''(cos(x)) = sin(x) cdot (- sin(x)) cdot (- sin(u)) + cos(x) cos(cos(x))`

`Rightarrow f''(cos(x)) = sin^(2)(x) cdot sin(u) + cos(x) cos(cos(x))`

Replace `u` with `cos(x)`:

`Rightarrow f''(cos(x)) = sin^(2)(x) cdot sin(cos(x)) + cos(x) cos(cos(x))`

`therefore f''(cos(x)) = sin^(2)(x) sin(cos(x)) + cos(x) cos(cos(x))`

Finally, let's evaluate `f''(cos(x))` at `x = frac(pi)(2)`:

`Rightarrow f''(cos(frac(pi)(2))) = sin^(2)(frac(pi)(2)) sin(cos(frac(pi)(2))) + cos(frac(pi)(2)) cos(cos(frac(pi)(2)))`

`Rightarrow f''(cos(frac(pi)(2))) = 1 cdot sin(0) + 0 cdot cos(0)`

`Rightarrow f''(cos(frac(pi)(2))) = 1 cdot 0 + 0 cdot 1`

`Rightarrow f''(cos(frac(pi)(2))) = 0 + 0`

`therefore f''(cos(frac(pi)(2))) = 0`

Therefore, the final answer is `0`.

Answer 2

`[f''(cos(x))]_(x=pi/2) = 0`

Explanation:

We have:

`f(x)=sinx`

And we want the second derivative of `f(cosx)`

Let:

`g(x) = f(cosx) = sin(cosx)`

Then by the chain rule we have:

`g'(x) = d/dx sin(cosx)`
`" " = (cos(cosx))(-sinx)`
`" " = -sinx \ cos(cosx)`

And for the second derivative we also need the product rule'

`g''(x) = d/dx -sinx \ cos(cosx)`
`" " = - \ d/dx sinx \ cos(cosx)`
`" " = - { (sinx)(d/dx cos(cosx)) + (d/dx sinx)(cos(cosx) }`

`" " = - { (sinx)(-sinx(cosx)(-sinx) + (cosx)(cos(cosx) }`

`" " = - sin^2x \ sinx(cosx) - cosx \ cos(cosx)`

And with `x=pi/2` we have:

`g(pi/2) = - sin^2(pi/2) \ sinx(cos(pi/2)) - cos(pi/2) \ cos(cos(pi/2))`

And as `sin(pi/2)=1`, `cos(pi/2)=0` and `sin0=0` we have

`g(pi/2) = - (1) \ sin(0) - (0) \ cos(0)`
`" " = - (1) \ (0)`
`" " = 0`