Question

How do you take the derivative of `tan(sqrt x)`?

Answer 1

`y^' = 1/2 * sec^2(sqrt(x))/sqrt(x)`

Explanation:

You can differentiate your function by using the chain rule for `y = tanu`, with `u = sqrt(x)`.

Let's assume that you don't know the derivative of `tanx`, but that you do know that

`d/dx(sinx) = cosx`

and

`d/dx(cosx) = -sinx`

Here's how you would determine `d/dx(tanx)`.

You know that `tanx = sinx/cosx`, which means that you can use the quotient rule to differentiate it

`d/dx(tanx) = ([d/dx(sinx)] * cosx - sinx * d/dx(cosx))/(cosx)^2`

`d/dx(tanx) = (cosx * cosx - sinx * (-sinx))/cos^2x`

`d/dx(tanx) = (overbrace(cos^2x + sin^2x)^(color(orange)(=1)))/cos^2x = 1/cos^2x = sec^2x`

Your target derivative will thus be

`d/dx(y) = d/(du) tanu * d/dx(u)`

`y^' = 1/cos^2u * d/dx(sqrt(x))`

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

`y^' = color(green)(1/2 * 1/sqrt(x) * 1/cos^2(sqrt(x)))`

or

`y^' = sec^2(sqrt(x)) * 1/2 * 1/sqrt(x)`

`y^' = color(green)(1/2 * sec^2(sqrt(x))/sqrt(x))`