Question

How do you differentiate `f(x)=1/cossqrt(1/x)` using the chain rule?

Answer 1

`-(sec(1/sqrt(x)) tan(1/sqrt(x)))/(2sqrt(x^3))`

Explanation:

The chain rule is used whenever there are functions of functions. In other words, when you have a function of the form;

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

The chain rule tells us that the derivative of this function is;

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

The first step to using the chain rule is to identify the separate functions. Lets start by rewriting the given function to make the component functions easier to identify.

`1/cos(sqrt(1/x)) = cos^-1((1/x)^(1/2))=cos^-1(x^(-1/2))`

Now we have a function of the form;

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

Where;

`f(x)=x^-1`
`g(x)=cosx`
`h(x)=x^(-1/2)`

You can see that if you work backwards plugging these functions into each other, you get the original function back.

`f(g(h(x)))=f(g(x^(-1/2)))) = f(cos(x^(-1/2))) = cos^-1(x^(-1/2))`

We can take the derivative of each of these component functions to get;

`f'(x)=-x^-2`
`g'(x)=-sinx`
`h'(x)=-x^(-3/2)/2`

Now we apply the chain rule to our function.

`d/(dx)cos^-1(x^(-1/2))=-cos^-2(x^(-1/2))d/dx cos(x^(-1/2))`

Apply the chain rule again.

`=-cos^-2(x^(-1/2))(-sin(x^(-1/2)))d/dx x^(-1/2)`

Take the last derivative.

`=-cos^-2(x^(-1/2))(-sin(x^(-1/2)))( -x^(-3/2)/2)`

Simplify.

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