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

1 Answer
Aug 13, 2015

#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))#