Question

How do you differentiate `f(x)=cos(sqrt((cosx^2)))` using the chain rule?

Answer 1

`f'(x) = (sin(sqrt(cos x^2)))/( sqrt(cos x^2)) cdot ( sin x^2)cdot x`

Explanation:

Use progressively the chain rule which says :
`(d F(u(x))) / dx = F'(u(x)) cdot u'(x)`.

`f'(x) = cos'(sqrt(cos x^2)) cdot (d sqrt(cos x^2))/dx`
so,
`f'(x) = -sin(sqrt(cos x^2)) cdot 1/(2 sqrt(cos x^2)) cdot (d cos x^2)/dx`
because `cos' = -sin` and `(d\sqrt(X))/(dX) = 1/(2sqrt(X))`.
Therefore;
`f'(x) = -(sin(sqrt(cos x^2)))/(2 sqrt(cos x^2)) cdot (-( sin x^2)cdot (d x^2)/(dx))`
Finally,
`f'(x) = (sin(sqrt(cos x^2)))/( sqrt(cos x^2)) cdot ( sin x^2)cdot x`
because `(d x^2)/(dx)=2x`.

Answer 2

`dy/dx = f^'(x) = (sin(sqrt(cos(x^2)))sin(x^2)x)/(sqrt(cos(x^2))`

Explanation:

I'll be using Leibniz notation because I think it's easier to understand more complicated chain rule applications with this.

The chain rule, in said notation goes as follows,

For `y = f(x)` such that `f(x) = g(u)` then `dy/dx = d/(du)(g(u))(du)/dx`,

Or, if we can say that a function `f(x)` is a simpler function `g(u)` for a defined relationship between `x` and `u`, we can derive `g(u)` wrt `u` and multiply by the derivative of `u`.

In this case we have

`y = cos(sqrt(cos(x^2)))`

By saying `sqrt(cos(x^2)) = u` we have

`dy/dx = d/(du)cos(u)(du)/dx`

Or

`dy/dx = -sin(u)(du)/dx`

Now, if we say that `cos(x^2) = v` we can say that

`dy/dx = -sin(u)d/(dv)(sqrt(v))(dv)/dx`

Or

`dy/dx = -sin(u)/(2sqrt(v))(dv)/dx = -sin(u)/(2u)(dv)/dx`

If we say `x^2 = w` we have

`dy/dx = -sin(u)/(2u)d/(dw)cos(w)(dw)/dx`

Or

`dy/dx = -sin(u)/(2u)(-sin(w))(dw)/dx = (sin(u)sin(w))/(2u)(dw)/dx`

Or, finally

`dy/dx = (sin(u)sin(w))/(2u)d/dx(x^2)`
`dy/dx = (sin(u)sin(w)x)/(u)`

And now we subsitute everything back to terms of `x`

`dy/dx = f^'(x) = (sin(sqrt(cos(x^2)))sin(x^2)x)/(sqrt(cos(x^2))`